MENSURATION 


AND 

PRACTICAL  GEOMETRY; 

CONTAINING 

TABLES  OF  WEIGHTS  AND  MEASURES,  VULGAR  AND 
DECIMAL  FRACTIONS, 

MENSUKATION  OF  AEEAS,  LIKES, 
SURFACES,  AND  SOLIDS, 

LENGTHS  OF  CIRCULAR  ARCS,  AREAS  OF  SEGMENTS  AND 

ZONES  OF  A  CIRCLE,  BOARD  AND  TIMBER  MEASURE, 

CENTRES  OF  GRAVITY,  &c,  &c. 

TO   WHICH  18   APPENDED   A 

TREATISE  ON  THE  CARPENTER'S  SLIDE-RULE  AND 

GAUGING. 


By  CHAS.  H.  HASWELL, 

CIVIL   AND   MAEINE   ENGINEER. 


SECOND  EDITION. 


OF  THE        ^>X\ 

SITY  ) 

y 

NEW    YORK: 

HARPER    &    BROTHERS,     PUBLISHERS, 

FRANKLIN    SQUARE. 

18  6  3." 


<Q.A4< 
W 


W 


a 


Entered,  according  to  Act  of  Congress,  in  the  year  one  thousand  eight  hundred 
and  sixty-three,  by 

HARPER  &  BROTHERS, 

In  the  Clerk's  Office  of  the  District  Court  of  the  Southern  District  of  New  York. 


TO 


B.  H.  BARTOL,  ESQ,, 


A  TRIBUTE  TO  AN  EARLY  AND  ESTEEMED  FRIEND  OF 


THE  AUTHOE. 

New  Yoek,  January*  1863. 


103910 


P  K  E  F  A  C  E. 


The  following  work  is  designed  for  the  use  of  Stu- 
dents, Mechanics,  and  Engineers,  and  the  purpose  of  the 
author  has  been  to  present  a  full  set  of  rules  whereby 
may  be  readily  determined  the  Lines,  Areas,  Surfaces, 
Solidities  or  Volumes,  and  Centres  of  Gravity  of  va- 
rious Regular  and  Irregular  Figures. 

In  the  progress  of  the  work,  it  was  essayed  to  pre- 
sent a  familiar  rule  for  the  surface,  volume,  and  centre 
of  gravity  of  every  figure,  but,  in  consequence  of  the  im- 
practicability of  reducing  the  rules  in  some  cases  to  a 
formula  at  all  consistent  with  the  design  of  the  work, 
such  cases  as  presented  this  difficulty  were  abandoned : 
as  it  occurs,  however,  the  number  of  them  has  been 
very  essentially  reduced  by  the  aid  and  advice  of  Pro- 
fessors A.  E.  Church,  U.  S.  Military  Academy,  West 
Point,  and  G.  B.  Docharty,  N.  Y.  Free  Academy. 

In  consequence  of  the  great  number  of  works  on 
Mensuration  that  have  been  submitted  to  the  public, 
the  inference  is  a  fair  one  that  the  author,  in  this  case, 
has  not  presented  any  novelties  whereby  he  may  antici- 
pate any  particular  notice  or  attention ;  he  trusts,  how- 
ever, that  a  reference  to  the  result  of  his  labors  will 
show  that,  in  the  essential  points  of  the  extent  of  the 
figures  submitted,  as  well  in  their  number  as  variety  of 


VI  PREFACE. 

section,  and  in  the  introduction  of  rules  for  determining 
their  centres  of  gravity,  he  has  submitted  some  features 
of  so  new  and  useful  a  purpose  as  to  entitle  him  to  the 
attention  of  those  upon  whom  he  confidently  relies  for 
patronage. 
New  York,  April  1st,  1858. 


CONTENTS, 


Page 

Explanation  op  Characters 9 

MEASURES  AND  WEIGHTS. 

Measures  of  Length 11 

Measures  of  Surface 12 

Measures  of  Capacity 14 

Measures  of  Solidity 15 

Measures  of  Weight 15 

Miscellaneous  Measures 16 

Measure  of  Value IT 

Ancient  Measures IT 

VULGAR  FRACTIONS.  19 

Reduction  of  Vulgar  Fractions 20 

Addition  of  Vulgar  Fractions 25 

Subtraction  of  Vulgar  Fractions 26 

Multiplication  of  Vulgar  Fractions. .  26 

Division  of  Vulgar  Fractions 2T 

Rule  of  Tliree  in  Vulgar  Fractions . .  2T 

DECIMALS.  28 

Addition  of  Decimals 28 

Subtraction  of  Decimals 29 

Multiplication  of  Decimals 29 

Division  of  Decimals 31 

Reduction  of  Decimals 32 

Rule  of  Three  in  Decimals S5 

DUODECIMALS 36 

Involution 3S 

Evolution 39 

To  extract  the  Square  Root 39 

Square  Roots  of  Vulgar  Fractions. . .  40 

To  extract  the  Cube  Root,  etc 41 

Properties  of  Numbers 44 

GEOMETRY 

Definitions 45 

MENSURATION  OF  AREAS,  LINES, 
AND  SURFACES. 

Parallelograms  (Square,  Rectangle, 
Rhombus,  Rhomboid,  and  Gno- 
mon)   4T-49 


Page 
Triangles 50 

Trapeziums  and  Trapezoids 5T 

Polygons 59 

Regular  Polygons 59 

Irregular  Figures 6T 

Circle 68 

Useful  Factors Tl 

Circular  Arc T2 

Section  of  a  Circle 81 

Segment  of  a  Circle 84 

Sphere 88 

Segment  of  a  Sphere 89 

Spherical  Zone 90 

Spheroids  or  Ellipsoids 91 

Circular  Zone 95 

Cylinder 93 

Prisms 99 

Wedges 100 

Trismoids 101 

Ungulas 102 

Lune 10T 

Cycloid 108 

Rings,  Circular 109 

Cylindrical 110 

Cones 112 

Pyramids 114 

Helix  (Screw). 116 

Spirals HT 

Spindles 120 

Circular  Spindle 120 

Elliptic,  Parabolic,  and  Hyperbolic . .  124 

Ellipsoid,  Paraboloid,  and  Hyper- 
boloid  op  Revolution 124 

Ellipsoid 125 

Paraboloid 126 

Any  Figure  op  Revolution 12T 


Vlll 


CONTENTS. 


Page 
Lengths  of  Circular  Arcs 130 

Areas  of  Segments  of  a  Circle 134 

Areas  of  Zones  of  a  Circle 139 

Promiscuous  Examples 144 

MENSURATION  OF  SOLIDS. 

Cube , 155 

Parallelopipedon 156 

Frisms,  Prismoids,  and  Wedges  . . .  156 

Regular  Bodies 160 

Tetrahedron 161 

Hexahedron,  Octahedron 162 

Dodecahedron,  Icosahedron 163 

Regular  Bodies 164 

Cylinder 165 

Cone 166 

Pyramid 168 

Spherical  Pyramid 171 

Cylindrical  Ungulas 1T1 

Sphere 176 

Segment  of  a  Sphere 177 

Spherical  Zone 178 

Spheroids  (Ellipsoids) 179 

Segments  of  Spheroids 180 

Frustrums  of  Spheroids '. .  1S2 

Cylindrical  Ring . 184 

Links,  Elongated  or  Elliptical 185 

Spherical  Sector 187 

Spindles 188 

Circular  Spindle 188 

Frustrum  or  Zone  of  Circular  Spindle  189 

Segment  of  Circular  Spindle 191 

Cycloidal  Spindle 192 

Elliptic  Spindle : 193 

Parabolic  Spindle 196 

Hyperbolic  Spindle 200 

Ellipsoid,  Paraboloid,  and  Hyper- 

boloid  of  Revolution  (Conoids). .  202 
Ellipsoid  of  Revolution  (Spheroid)  . .  203 


Pag« 

Paraboloid  of  Revolution 203 

Hyperboloid  of  Revolution 206 

Any  Figure  of  Revolution ....  209 

Irregular  Body  .'. 210 

Promiscuous  Examples 211 

CONIC  SECTIONS. 

Definitions 228 

Conoids 233 

Ellipse 234 

Parabola 240 

Parabolic  Curve 24? 

Area  of  Parabola 244 

Hyperbola 246 

Arc  of  Hyperbola 250 

Conic  Ungulas 253 

Regular  Polygons 258 

Table  of  Units  for  Elements 261 

Mensuration  of  Regular  Bodies  or 

Polyhedrons 262 

Tables  of  Units  for  Elements 267 

Orophoids  : 

Domes,  Arched  and  Vaulted  Roofs. . .  268 

Saloons 269 

Board  and  Timber  Measure 276 

APPENDIX. 

Mensuration  of  Surfaces 278 

Centres  of  Gravity  of  Surfaces  . .  281 

Mensuration  of  Solids 285 

Solids  of  Revolution 286 

Centres  of  Gravity  of  Solids 292 

Sjjherical  Triangles,  Pyramids,  Epi- 
cycloids, Heli-coids,  &c,  &c 296 

Carpenters'  Slide-Rule 297 

Gauging 307 

Ullaging  of  Casks 319 


EXPLANATIONS    OF   CHARACTERS 

Used  in  the  following  Calculations,  etc.,  etc. 


Equal  to,  as  12  inches=l  foot,  or  8x8  =  16x4. 

Plus,  or  more,  signifies  addition;  as  4  +  6+5  =  15. 

Minus,  or  less,  signifies  subtraction;  as  15  —  5  =  10. 

Multiplied  by,  or  into,  signifies  multiplication ;  as 
8x9=72.  axd,  a.d,  or  ad,  also  signify  that  a 
is  to  be  multiplied  by  d. 

Divided  by,  signifies  division;  as  72-j-9=8. 

Is  to,  I 

o    •     I  signifies  Proportion,  as  2  :  4::8  :  16;  that 

'  f      is,  as  2  is  to  4,  so  is  8  to  16. 
To,     J 

V  Radical  sign,  which,  prefixed  to  any  number,  sig- 

nifies that  the  square  root  of  that  number  is  re- 
quired; as  V9,  or  Vaxb.  The  degree  of  the 
root  is  indicated  by  the  number  placed  over  the 
sign,  which  is  termed  the  index  of  the  root,  or 
radial;  as  ^,  y ,  etc. 

',   *,  added  or  set  superior  to  a  number,  signifies  that 

that  number  is  to  be  squared,  cubed,  etc. ;  thus  43 
means  that  4  is  to  be  multiplied  by  4 ;  43,  that  it 
is  to  be  cubed,  as  43  is  =4  X  4  X  4  =  64.  The  pow- 
er, or  number  of  times  a  number  is  to  be  multi- 
plied by  itself,  is  shown  by  the  number  added,  as 

3     3     4     5     pfp 

The  vinculum,  or  bar,  signifies  that  the  numbers  are 
to  be  taken  together;  as  8  —  2  +  6=12,  or 
3x5+3=24. 

Decimal  point,  signifies,  when  prefixed  to  a  number, 
that  that  number  has  a  unit  (1)  for  its  denom- 
inator; as  .1  is  •jJtf,  .155  is  i^j-,  etc. 

co  Difference,   signifies,  when    placed  between   two 

quantities,  that  their  difference  is  to  be  taken. 


X  EXPLANATIONS   OF   CHARACTERS. 

Signify  Degrees,  minutes,  seconds,  and  thirds. 

/_  Signifies  angle. 

±  Signifies  perpendicular. 

A  Signifies  triangle, 

D  Signifies  square,  as  □  inches. 

>     T  Signify  inequality,  or  greater  than,  and  are  pnt  be- 

tween two  quantities ;  as  a  T  b  reads  a  greater 
than  b. 

<      L  Signify  the  reverse ;  as  a  lb  reads  a  less  than  b. 

Signifies  therefore  or  hence. 
Signifies  because. 

(  )    [.]  Parentheses  and  brackets  signify  that  all  the  fig- 

ures within  them  are  to  be  operated  upon  as 
if  they  were  only  one ;  thus  (3  +  2)  X  5  =  25 ; 
[3 +2]  X  5  =25. 

p  or  it  Is  used  to  express  the  ratio  of  the  circumference  of 

a  circle  to  its  diameter  =  3.1415926,  etc. 
A  A'  A"  A'"     Signify  A,  A  prime,  A  second,  A  third,  etc. 

±     tf  Signify  that  the  formula  is  to  be  adapted  to  two 

distinct  cases. 

a~ ',  a-2,  a~3,  etc.  Denote  inverse  powers  of  a,  and  are  equal  to 

ill 

S*  a*  a*  etC* 

sin.^a,  cos.~*a,  etc.  Signify  the  arc  or  angle,  the  sine  or  cosine,  etc.,  of 
which  is  a,  the  arc  or  angle  being  expressed  in 
terms  of  the  radius,  as  the  unit,  unless  other- 
wise stated.     If  A0  denotes  the  arc  or  angle  in 

•7rA° 

degrees,  in  terms  of  the  radius  it  is  A=— — . 

If  sin.  A.— a,  ,'.sin.  ~~  1a=A,  etc. 

,    ,  etc.  Set  superior  to  a  number,  signify  the  square  or 

cube  root,  etc.,  of  the  number;  as  22  signifies 
the  square  root  of  2. 

I  #  ?- 

a,  3,  a,  etc.        Set  superior  to  a  number,  signify  the  square  or  cube 

root,  etc.,  of  the  4th  power,  etc.,  etc. 

*-*\  3*6,  etc.        Set  superior  to  a  number,  signify  the  tenth  root  of 

the  17th  power,  etc.,  etc. 


'**       OF  THE    ^4fa 

UNIVERSITY 


HASWELL'S  MENSURATION, 


MEASURES  AND  WEIGHTS 

Used  in  this  Work, 
MEASURES  OF  LENGTH. 

LINEAL. 


12  inches  =  1  foot. 
3  feet  =  1  yard. 
5  J  yards   =  1  rod  or  pole. 


40  rods       —1  furlong. 
8  furlongs =1  mile. 
1  degree    =69.77  statute 
miles. 
1  geographical  mile=2046.58  yards  or  6139.74  feet. 


CIRCLES. 

60  thirds    =1  second.  i     60  minutes =1  degree. 

60  seconds  =1  minute.  |  360  degrees  =1  circle. 

1  day  is       .002739  of  a  year. 

1  minute  is  .000694  of  a  year. 


6  points =1  line. 
12  lines   =1  inch. 
1  palm  =3  inches. 


MISCELLANEOUS. 

1  hand =4  inches. 
1  span  ss  9  inches. 
6  feet   =1  fathom. 
1  yard =.000568    of  a  mile. 
1  foot  =.000199  " 

1  inch  =.0000158        " 
Gunter's  Chain  is  4  poles  or  22  yards  in  length,  and  has 
100  equal  links  of  .666  of  a  foot,  or  7.92  inches. 
80  chains=l  mile. 


12 


MEASURES    AND   WEIGHTS. 


Foreign. 


Great  Britain. 

Imperial  yard 

=39.1393  imperial  inches. 

Mile 

=  1760  U.  S.  yards. 

France. 

Metre 

=39.37079  inches,  or  3.2809  feet 

Foot  (old  system) 

=  12.7925  inches. 

Common  league 

=4264.16  U.  S.  yards. 

Austria. 

Foot 

=  12.445  inches.* 

Mile 

=8296.66  U.  S.  yards. 

China. 

Foot,  builder's 

=  12.71  inches. 

"     mathematic 

=  13.12      " 

"     surveyor's 

=  12.58      " 

"     tradesman's 

=  13.32      " 

Li 

=629  U.  S.  yards. 

Copenhagen. 

Foot 

=  12.35  inches. 

Genoa. 

Foot 

=9.72       " 

Hamburgh. 

Foot 

=  11.29     " 

Mile 

=8244  U.  S.  yards. 

Lisbon. 

Foot 

=  12.96  inches. 

Mexico. 

Foot 

=  11.1284  inches. 

Common  league 

=4636.83  U.  S.  yards. 

Prussia. 

Foot 

=  12.361  inches. 

Mile 

=8468  U.  S.  yards. 

Rome." 

Foot 

=  11.60  inches. 

Mile 

=2025  U.  S.  yards. 

Russia. 

Foot 

=21.1874  inches.* 

Versta 

=  1167  U.  S.  yards. 

Spain. 

Foot 

=  11.1284  inches.*        ■      ' 

Judicial  league 

=4636.83  U.  S.  yards. 

Sweden. 

Foot 

=  11.6865  inches.* 

Mile 

=  11700  U.  S.  yards. 

Turkey. 

Pick 

=  17.905  inches. 

Berri 

=1826  U.  S.  yards. 

MEASURES  OF  SURFACE. 


144    inches    =1  foot. 

9    feet        =1  yard 

272ifeet    "> 


SQUARE. 

40  rods  =1  rood. 

4  roods =1  acre. 

640  acres  =  1  mile. 


SOi  yards) 


—  1  rod  or  pole. 


*  U.  S.  Ordnance  Manual,  1850. 


WEIGHTS    AND    MEASURES. 


13 


10  square  chains 
4840       "      yards 
160       "      poles 
100000       "       links 
69.5701  yards  square 
220x198  feet 
208.710321  feet  square 
235.5041  feet  diameter,  or 
43560  square  feet 


►  =  1  acre. 


MISCELLANEOUS. 

24  sheets^l  quire.  20  quires  =  1  ream. 


Drawing  Paper. 


Cap 13    xl6 

Demy 19Jxl5J 

Medium 22    xl8 

Koyal 24    xl9 

Super-royal  ...  27    X 19 

Imperial 29    x21J 

Elephant 27|x22i 


in. 
u 


Columbier 33| 

Atlas 33 

Doub.  Elephant  40 
Theorem  .....  34 
Antiquarian. . .  52 

Emperor 40 

Uncle  Sam ....  48 


X 

23  in 

x 

26  " 

X 

26  " 

X 

28  " 

X 

31  " 

X 

60  " 

Xl20 


Foreign. , 


France.     Old  System, 

New  System,  1  are= 
Amsterdam. 
Berlin. 
Hamburgh. 
Portugal. 
Prussia. 
Rome. 
Russia. 
Spain. 
Switzerland. 


1  square  inch  =         1.13587  U.  S.  inch. 

:  100  square  metres  =     1 19.603  square  yards. 

Morgen  =  9722  "  " 

*        great      =  6786  "  " 

"  =11545  "  " 

Geira  =  6970  "  " 

Morgen  =  3053  "  " 

Pezza  ==  3158  "  " 

Desiatina  =13066.6  "  " 

Fanegada  =  5500  "  .       " 

Taux  =  7855  "  " 


Square  Foot  in  U.  S.  Square  Inches. 


Amsterdam 124.255 

Antwerp 126.337 

Berlin 148.603 

Bologna 224.700 


Bremen 129.504 

Cologne 117.288 

Dantzic 127.690 

Denmark 152.670 


14 


MEASURES    AND   WEIGHTS. 


Dresden 124.099 

France 163.558 

Geneva 369.024 

Hamburgh  127.441 

Leipsic 123.432 

Lisbon 167.547 

Milan 243.984 


Rhineland 152.670 

Riga 116.425 

Rome 137.358 

Spain 123.832 

Sweden 136.515 

Venice 187.182 

Vienna 155.002 


MEASURES  OF  CAPACITY 

LIQUID   AND   DRY. 


7.21875  cub.  i 
4  gills 

2  pints 


1  gill- 

1  pint. 
1  quart. 


4  quarts  =  1  gallon. 
2  gallons  =1  peck. 
4  pecks    ss  1  bushel. 


MISCELLANEOUS. 

1  chaldron— 36  bushels,  or  57.244  cubic  feet,  when  heaped 
in  the  form  of  a  cone. 

Note. — The  standard  U.  S.  bushel  is  the  Winchester  (British),  and  it 
measures  2150.42  cubic  inches,  and  contains  543391.89  troy  grains,  or 
77.627413  pounds  avoirdupois  of  distilled  water  at  its  maximum  density. 

Its  dimensions  are  18£  inches  in  diameter  inside,  19£  inches  outside, 
and  8  inches  deep.  When  heaped,  the  cone  must  not  be  less  than  6 
inches  high,  and  it  contains  2986.4765  cubic  inches. 

The  standard  U.  S.  gallon=231  cubic  inches,  and  contains  58372.1754 
troy  grains  (8.3389  pounds  avoirdupois)  of  distilled  water  at  its  maxi- 
mum density  (39°.83). 


Foreign. 


Great  Britain. 


France/ 


Amsterdam. 
Antwerp. 
Bremen. 
Constantinople. 


The  imperial  gallon  measures  277.274  cubic  inches. 
Imperial  bushel  2218.192  cubic  inches,  and  when 
neaped  in  the  form  of  a  true  cone  (6  inches  high) 
it  contains  2815.4872  cubic  inches. 
1  chaldron =58.68  cubic  feet,  and  weighs  3136  lbs. 
Old  System.  1  Pinte=0.931  litre,  or  56.817  cubic 

inches. 
New  System.  1  Litre =61.027  U.  S.  inches. 
Anker         2331    cub.  in.       Mudde       6786  cub.  in. 
Stoop  168  "  Viertel       4705      " 

Stubgens      194.5       "  Scheffel      4339      " 

Almud  319  "  Kislos        2023      " 


MEASURES    AND    WEIGHTS. 


15 


Copenhagen. 

Anker 

2335    cub.  in. 

Toende 

8489  cub.  in. 

Genoa. 

Pinte 

90.5       " 

Mina 

7366      " 

Lisbon. 

Almudi 

1010 

Alqueire 

827      " 

Rome. 

Boccali 

80 

Quarti 

4226      " 

Russia. 

Vedro 

750.58     " 

Chet.wert  12800      " 

Spain. 

Quartillos 

30.5       " 

Catrize 

41269      " 

■ 

Arroba 

4.2455  gall. 

Fanega 

1.593  bush. 

Sweden. 

Kann 

160    cub.  in. 

Tunnar 

8940  cub.  in. 

Tripoli. 

Mattari 

1376          " 

"  Caffiri 

19780      " 

Vienna. 

Eimer 

3443          « 

Metzen 

3753      " 

MEASURES  OF  SOLIDITY. 

CUBIC. 

1728  inches=l  foot.  |  27  feet  =  l  yard. 

1       foot =7.4806  gallons. 
128       feet=:l  cord. 

24.75    "  =1  perch. 


Foreign. 

France.     Stere  (1  cubic  metre) =6 102 7.1  U.  S.  inches,  or 
35.3166  cubic  feet. 

Note. — For  the  solid  measures  of  other  foreign  countries,  take  the 
cube  of  the  measures  given  in  the  preceding  tables. 


MEASUBES  OF  WEIGHT. 


AVOIRDUPOIS. 


1G  drams  =1  ounce. 
1G  ounces =1  pound. 


112  pounds  =1  cwt. 
20  cwt.      =1  ton. 


TROY  WEIGHT. 

24  grains  r=l  pennyweight. 

20  penny  weights  =  1  ounce. 
12  ounces  =1  pound. 


16 


MEASURES    AND    WEIGHTS. 


The  pound,  ounce,  and  grain  are  the  same  in  apothecaries' 
and  troy  weights. 

7000  troy  grains  =  1 
175  «  pounds=144 
175    "     ounces  =192 


lb.  avoirdupois, 
lbs.         " 
oz.  " 

oz.  " 

.8228  lb.  " 

The  standard  U.  S.  pound  contains  7000  troy  grains,  or 
27.7015  cubic  inches  of  distilled  water  at  its  maximum 
density. 


437.5  troy  grains  =     1 
1        "     pound  = 


Foreign. 


Great  Britain. 

Pound  avoi 

irdupois 

=27.7274  cubic 

:  inches  of  ( 

tilled  water  at  the  temperature  < 

of  62°.     Hen 

22.815689  cubic  inches  weigh  1  troy  pound. 

France. 

1  Gramme 

= 

15.43316 

troy  grains. 

Alexandria. 

1  Rottoli 

5= 

.9346 

pounds  avoirdupois. 

Amsterdam. 

1  Pound 

= 

1.0893 

a 

(i 

Austria. 

1      " 

BS 

1.2351 

u 

H 

Bengal. 

1  Seer 

55 

1.8667 

" 

M 

Bremen. 

1  Pound 

BS 

1.0997 

u 

M 

Cairo. 

1  Rottoli 

= 

.9523 

ii 

U 

China. 

1  Catty 

= 

1.3253 

« 

11 

Constantinople. 

1  Oke 

BS 

2.8129 

(I 

(I 

Copenhagen. 

1  Pound 

BT 

1.1014 

u 

U 

Corsica. 

1      " 

SB 

.7591 

II 

u 

Genoa. 

1      "       (heavy)  == 

1.0768 

U 

a 

Japan. 

1  Catty 

BS 

1.3000 

a 

(< 

Prussia. 

1  Pound 

BE 

1.0333 

u 

II 

Rome.  I 

1      " 

as 

.7479 

U 

ii 

Russia. 

1      " 

S3 

.9020 

II 

ii 

Spain. 

1      "     • 

= 

1.0152 

(4 

it 

Sweden. 

1      " 

SB 

.9376 

II 

ii 

Tripoli. 

1  Rottoli 

BS 

1.1200 

II 

it 

Venice. 

1  Pound  (heavy)  = 

1.0555 

M 

it 

MISCELLANEOUS. 

1  cubic  foot  of  anthracite  coal  from  50  to  55  lbs. 
1  cubic  foot  of  bituminous  coal  from  45  to  55  lbs. 
1  cubic  foot  of  Cumberland  coalm     53     lbs. 


MEASURES    AND    WEIGHTS. 


17 


1  cubic  foot  of  charcoal   =     18.5  lbs.  (hard  wood.) 
1  cubic  foot  of  charcoal    =     18.      "    (pine  wood.) 
1  cord  Virginia  pine         =2700       " 
1  cord  Southern  pine       =3300       " 
1  stone  =     14       " 

Coals  are  usually  purchased  at  the  conventional  rate  of  28 
bushels  (5  pecks)  to  a  ton =43.56  cubic  feet. 

MEASURES  OF  VALUE. 

1  eagle  =258     troy  grains. 
1  dollar=412.5    «         " 
1  cent    =168       "         " 
The  standard  of  gold  and  silver  is  900  parts  of  pure  metal 
and  100  of  alloy  in  1000  parts  of  coin. 


A  digit 
A  palm 
A  span 


ANCIENT  MEASURES. 

MEASURES    OF   LENGTH. 


Scripture. 


0.912 

3.648 

10.944 


A  cubit         =1 
A  fathom     =7 


9.888 
3.552 


Feet 

=0 


Grecian. 


A  digit  =0     0.7554^-        A  stadium: 

A  pous  (foot)=l     0.0875  A  mile 

A  cubit  =1     1.5984| 

A  Greek  or  Olympic  foot= 12.108  inches. 

A  Pythic  or  natural  foot  =   9.768       " 


Feet. 

604 
4835 


Inches. 

4.5 


Jewish. 


A  cubit 
A  Sabbath  day's 
journey 


=        1.824 


=3648. 


A  mile  =     7296 

A  day's  journey  =175104 

(or  33  miles  864  feet). 


18 


MEASURES    AND   WEIGHTS. 


Roman. 


A  digit  : 

An  uncia  (inch): 
A  pes  (foot) 


Arabian  foot 
Babylonian  foot 
Egyptian        " 


Inches. 

.72575 
.967 
11.604 


Feet.  Inches. 

A  cubit  =        1  5.406 

A  passus=       4  10.02 
A  mile    =4835 


Miscellaneous. 


Feet. 

=  1.095 
=  1.140 
=  1.421 


Feet. 

Hebrew  foot  =1.212 

"        cubit  =1.817 

"        sacred  cubit= 2.002 


FRACTIONS.  19 


VULGAE  FRACTIONS. 

A  Fraction,  or  broken  number,  is  one  or  more  parts  of 
a  Unit. 

Illustration. — 12  inches  are  1  foot. 

Here  1  foot  is  the  unit,  and  12  inches  its  parts ;  3  inches, 
therefore,  are  the  one  fourth  of  a  foot,  for  three  is  the  quarter 
or  fourth  of  12. 

A  Vulgar  Fraction  is  a  fraction  expressed  by  two  numbers 
placed  one  above  the  other,  with  a  line  between  them,  as  50 
cents  is  the  \  of  a  dollar. 

The  upper  number  is  called  the  Numerator,  because  it  shows 
the  number  of  parts  used. 

The  lower  number  is  called  the  Denominator,  because  it  de- 
nominates, or  gives  name  to  the  fraction. 

The  Terms  of  a  fraction  express  both  numerator  and  de- 
nominator ;  as  6  and  9  are  the  terms  of  f . 

A  Proper  fraction  has  the  numerator  equal  to,  or  less  than 
the  denominator ;  as  \,  -J,  &c. 

An  Improper  fraction  is  the  reverse  of  a  proper  one ;  as 
f ,  4,  &c. 

A  Mixed  fraction  is  a  compound  of  a  whole  number  and  a 
fraction ;  as  5-J ,  &c. 

A  Compound  fraction  is  the  fraction  of  a  fraction ;  as  \  of 

f,4  of-!,  &c. 

A  Complex  fraction  is  one  that  has  a  fraction  for  its  numer- 

i        3— 
ator  or  denominator,  or  both ;  as  i,  or  %,  or  J,  or  S,  &c. 

A  Fraction  denotes  division,  and  its  value  is  equal  to  the  quotient 
obtained  by  dividing  the  numerator  by  the  denominator  ;  thus  ±£-  is 
equal  to  3,  -^  is  equal  to  44,  and  %  is  equal  to  -^. 


20  FRACTIONS. 


REDUCTION  OF  VULGAR  FRACTIONS. 

To  find  the  greatest  Number  that  will  divide  two  or 
more  Numbers  without  a  Remainder, 

Rule. — Divide  the  greater  number  by  the  less ;  then  divide 
the  divisor  by  the  remainder ;  and  so  on,  dividing  always  the 
last  divisor  by  the  last  remainder,  until  nothing  remains. 

When  there  are  more  than  two  numbers,  find  the  greatest 
common  measure  of  two  of  them,  and  then  that  for  this  com- 
mon measure  and  the  remaining  number. 

Example. — What  is  the  greatest  common  measure  of  1908 
and  936?  936)1908(2 

1872 

36)936(26 

72 

216 

216 

Hence  36  is  the  greatest  common  measure. 

Ex.  2.  What  is  the  greatest  common  measure  of  246  and 

372  ?  Ans.  6. 

Ex.  3.  What  is  the  greatest  common  measure  of  1728,  864, 
and  3456  * 

864)3456(4 
3456 
864  is  the  greatest  common  measure  of  3456  and  864. 
864)864(1 
864 
Hence  864  is  the  greatest  common  measure  of  the  three  num- 
bers. 

Ex.  4.  What  is  the  greatest  common  measure  of  216  and 
288?  AnsM2. 

To  find  the  least  common  Multiple  of  two  or  more 
Numbers., 

Rule. — Divide  by  any  number  that  will  divide  two  or  more 


FRACTIONS.  21 

of  the  given  numbers  without  a  remainder,  and  set  the  quo- 
tients with  the  undivided  numbers  in  a  line  beneath. 

Divide  the  second  line  as  before,  and  so  on,  until  there  are  no 
two  numbers  that  can  be  divided ;  then  the  continued  product 
of  the  divisors  and  quotients  will  give  the  multiple  required. 
Example. — What  is  the  least  common  multiple  of  40,  50, 
and  25? 

5)40  .  50 .  25 

5)  8.10.    5 

2)  8.    2.    1 

4.    1.    1 

Then  5x5x2  x  4=200,  Ans. 

To  reduce  Fractions  to  their  lowest  'Terms. 

Rule. — Divide  the  terms  by  any  number  that  will  divide 
them  without  a  remainder,  or  by  their  greatest  common  meas- 
ure at  once. 

Example. — Reduce  -£§ $  of  a  foot  to  its  lowest  terms. 
l™  +  10=™  +  8=^+3=%,  or  9  inches. 
.  Ex.  2.  Reduce  -|ff  to  its  lowest  terms.  Ans.  § . 

To  reduce  a  Mixed  Fraction  to  its  equivalent  Improper 
Fraction. 

Note. — Mixed  and  improper  fractions  are  the  same;  thus  5£=-2J-. 

Rule. — Multiply  the  whole  number  by  the  denominator  of 
the  fraction,  and  \g  the  product  add  the  numerator,  then  set 
that  sum  above  the  denominator. 

Example. — Reduce  23J  to  a  fraction. 

23x6  +  2     140     ,i 

-— ■ — = — — = the  answer. 

o  6 

Ex.  2.  Reduce  20$  inches  to  a  fraction.  Ans.  ±%2-. 

Ex.  3.  Reduce  5-J  to  a  fraction.  Ans.  $£-. 

Ex.  4.  Reduce  183^5T  to  a  fraction.  Ans.  3|j9. 

Ex.  5.  Reduce  125£  to  a  fraction.  Ans.  £JI* 


FRACTIONS. 


To  reduce  an  Improper  Fraction  to  its  equivalent 
Whole  or  Mixed  Number. 

Rule. — Divide  the  numerator  by  the  denominator,  and  the 
quotient  will  be  the  whole  or  mixed  number  required. 
Example. — Reduce  ^  to  its  equivalent  number. 

^  or  12-^-3=4,  the  answer. 
Ex.  2.  Reduce  ^^  to  its  equivalent  number.       Ans.  8. 

To  reduce  a  Whole  Number  to  an  equivalent  Fraction 
having  a  given  Denominator. 

Rule. — Multiply  the  whole  number  by  the  given  denomi- 
nator, and  set  the  product  over  the  said  denominator. 

Example. — Reduce  8  to  a  fraction  whose  denominator  shall 
be  9. 

8  x  9=72 ;  then  ^-—the  answer. 

Ex.  2.  Reduce  12  to  a  fraction  whose  denominator  shall 
be  13.  Ans.  J#, 

To  reduce  a  Compound  Fraction  to  an  equivalent 
Simple  one. 

Rule. — Multiply  all  the  numerators  together  for  a  numer- 
ator, and  all  the  denominators  together  for  a  denominator. 
N0tE# —  When  there  are  terms  that  are  common,  they  may  be  omitted. 
Example. — Reduce  \  of  f  of  §  to  a  simple  fraction. 
ix|x#=A=i,  Am. 
Or,  -  x  -X-=-»  ty  canceling  the  2's  and  3's. 

Ex.  2.  Reduce  -|  of  f  to  a  simple  fraction. 

ix|=f,  Ans. 
Ex.  3.  Reduce  £  of  %  to  a  simple  fraction.  Ans.  Jf. 

Ex.  4.  Reduce  •§  of  f  of  -f  of  £  of  -^  to  a  simple  fraction. 

Ans.  *%. 
Ex.  5.  Reduce  2,  and  -|  of  £  to  a  fraction.  ^4tw.  §#. 

Ex.  6.  Reduce  2^  and  $  of  §  to  a  fraction.  -4ws.  £. 


FRACTIONS.  23 

To  reduce  Fractions  of  different  Denominations  to 
equivalent  ones  having  a  common  Denominator. 

JJule. — Multiply  each  numerator  by  all  the  denominators 
except  its  own  for  the  new  numerators,  and  multiply  all  the 
denominators  together  for  a  common  denominator. 

Note. — In  this,  as  in  all  other  operations,  whole  numbers,  mixed,  or  com- 
pound fractions,  mustjirst  be  reduced  to  the  form  of  simple  fractions. 

Example. — Keduce  -J,  -§-,  and  f  to  a  common  denominator. 
1x3x4  =  12^ 


2x2x4  =  16 
3x2x3  =  18 


12     16     18      An<! 
Til  -2TJ  Tfj  JinS' 


2x3x4  =  24^ 
The  operation  may  be  performed  mentally  ;  thus, 
Reduce  -J-,  -§>  !>  an(^  "i  to  a  common  denominator. 

1  1  3 12  6 6     n71fJ  5 2  0 

"5"  —  "§"*  H  —  ~W~'  "F  —  ~S>  ana  ~2  —  -W 

Ex.  2.  Reduce  2,  and  -§  of  -J  to  a  common  denominator. 

A„o    60     12     25 

Ex.  3.  Reduce  -§ ,  2f ,  and  4  to  a  common  denominator. 

Ana     25      78      120 
XX /to.   ^-y-,  -g-g,  -Jo-* 

Ex.  4.  Reduce  4,  ^,  14,  and  54  to  a  common  denominator. 


-2~>  4>  A"2">  *""  WB~ 


J.,,     24      36      72      256 
AB*   4¥?  T¥J   4~g->  _4~r* 


Note  1.  When  the  denominators  of  two  given  fractions  have  a  com- 
mon measure,  divide  them  by  it ;  then  multiply  the  terms  of  each  given 
fraction  by  the  quotient  arising  from  the  other  denominator. 

Ex.  5.  Reduce  ■£$  and  -^  to  a  common  denominator. 

^andif=^and^,Ans. 

Note  2.  When  the  less  denominator  of  two  fractions  exactly  divides 
the  greater,  multiply  the  terms  of  that  which  has  the  less  denominator 
by  the  quotient. 

Ex.  6.  Reduce  -f-  and  -^  to  a  common  denominator. 

5 

f  and  1?=-^  and  ^,  Ans. 

To  reduce  Complex  Fractions  to  Simple  ones. 

Rule. — Reduce  the  two  parts  both  to  simple  fractions,  then 
multiply  the  numerator  of  each  by  the  denominator  of  the  other. 


24  FRACTIONS. 


22 
Example. — Simplify  the  complex  fraction  jj. 


2|=f  8x    5=40_fl    A 

Ex.  2.  Simplify  the  complex  fraction  f. 


|=4     f==fc  Am- 

8f 

Ex.  3.  Simplify  the  complex  fraction  jf.  ^4w5.  -Jf* 

7b  ^/mtZ  £/i<?  Value  of  a  Fraction  in  parts  of  a  whole 
Number. 

Rule. — Multiply  the  whole  number  by  the  numerator,  and 
divide  by  the  denominator ;  then,  if  any  thing  remains,  mul- 
tiply it  by  the  parts  in  the  next  inferior  denomination,  and 
divide  by  the  denominator  as  before,  and  so  on  as  far  as  nec- 
essary ;  so  shall  the  quotients  placed  in  order  be  the  value  of 
the  fractions  required. 

Example. — What  is  the  value  of  J  of  ■§•  of  $9  ? 
J  of  § =£,  and  f  of  £=J^-=$3,  Ans. 
Ex.  2.  Reduce  J  of  a  pound  to  avoirdupois  ounces. 
3 
1 
4)  3(0  lbs. 

16  ounces  in  a  lb. 
4)48 
12  ounces,  Ans. 
Ex.  3.  Reduce  -^  of  a  day  to  hours. 

i^X^=li=7£>  hours,  Ans. 
Ex.  4.  Reduce  %  of  a  pound  troy  to  ounces  and  pennyweights. 
4 
12 
5)48 
ounces  9     3 
20 
5)60 
pennyweights  12=9  oz.,  12  dwts.,  Ans. 


FRACTIONS.  25 

Ex.  5.  What  is  the  value  of  $  of  an  acre  ? 

.Arcs.  3  roods,  20  poles. 
Ex.  6.  What  is  the  value  of  f  of  $4  83  ? 

Ans.  $1  93J. 

2b  reduce  a  Fraction  from  one  Denomination  to 
another. 

Rule. — Multiply  the  number  of  parts  in  the  next  less  de- 
nomination by  the  numerator  if  the  reduction  is  to  he  to  a  less 
name,  but  multiply  by  the  denominator  if  to  a  greater. 

Example. — Reduce  J  of  a  dollar  to  the  fraction  of  a  cent. 

Ex.  2.  Reduce  \  of  an  avoirdupois  pound  to  the  fraction 
of  an  ounce. 

is/  16 16 8      An* 

Ex.  3.  Reduce  -f-  of  a  cwt.  to  the  fraction  of  a  lb. 

2  VH2 224 32       A„Q 

Ex.  4.  Reduce  §  of  f  of  a  mile  to  the  fraction  of  a  foot. 

2   nf  3 6    w  6280 31680 2640      A^o 

~s  0I  ¥— T2  x — i — — — ra — — — i — J  ^ns. 
Ex.  5.  Reduce  J  of  a  square  foot  to  the  fraction  of  an  inch. 

Ans.  -2j£. 

ADDITION  OF  VULGAR  FRACTIONS. 

Rule. — If  the  fractions  have  a  common  denominator,  add 
all  the  numerators  together,  and  place  their  sum  over  the  de- 
nominator. 

Note. — If  the  fractions  have  not  a  common  denominator,  they  must  be 
reduced  to  one,  and  compound  and  complex  must  be  reduced  to  simple  frac- 
tions. 

Example. — Add  \  and  J  together. 

i+4=|=l,  Ans. 
Ex.  2.  Add  J  of  J  of  tV  to  2f  of  f . 

Then,  W+«=-,ftW+MS*=l«*4=144fc  Ans' 

Ex.  3.  Add  |  and  $  together.  Ans.  1^%. 

Ex.  4.  Add  I,  7fc  and  }  of  |  together.  Ans.  8|. 

B 


26  FRACTIONS. 

/ 

Ex.  5.  Add  T  of  an  eagle,  -J  of  a  dollar,  and  -j^  of  a  cent 
together.  Ans.  165f|f. 

SUBTRACTION  OF  VULGAR  FRACTIONS. 

Rule. — Prepare  the  fractions,  when  necessary,  the  same  as 
for  other  operations,  then  subtract  the  one  numerator  from 
the  other,  and  set  the  remainder  over  the  common  denom- 
inator. 

Example. — What  is  the  difference  between  -f  and  ^  ? 

|-i=|»  Ans. 
Ex.  2.  Subtract  -§  from  -§. 
6x9=54\ 

3x8  =  24  Uff-ff=ff,  Ana. 
8x9  =  72j 
Ex.  3.  Subtract  -^  from  ^-.  Ans.  T^-. 

Ex.  4.  Subtract  $  of  -^  from  f  of  5  J-  of  1.      ,4ns.  -|f$£. 


MULTIPLICATION  OF  VULGAR  FRACTIONS. 

Rule. — Prepare  the  fractions,  when  necessary,  as  previous- 
ly required ;  multiply  all  the  numerators  together  for  a  new 
numerator,  and  all  the  denominators  together  for  a  new  de- 
nominator. 

Example. — What  is  the  product  of  -§  and  §  ? 

ix%=&=ih,Ans. 
Ex.  2.  What  is  the  product  of  6  and  %  of  5  ? 
6xf  of5=6x-^=-$f=20,  Ans. 
Ex.  3.  What  is  the  product  of  § ,  3£,  5,  and  f  of  f  ? 
3 


-x^x^x(fof^=^xf=¥=41,^. 

2  4 

Ex.  4.  What  is  the  product  of  5,  f,  f  of  %  and  4£? 


FRACTIONS.  27 


DIVISION  OF  VULGAR  FRACTIONS. 

Rule. — Prepare  the  fractions,  when  necessary,  as  previous- 
ly required ;  then  divide  the  numerator  by  the  numerator,  and 
the  denominator  by  the  denominator,  if  they  will  exactly  di- 
vide ;  but  if  not,  invert  the  terms  of  the  divisor,  and  multiply 
the  dividend  by  it,  as  in  multiplication. 

Example. — Divide  -%§.  by  f . 

25    .    5 5 12       Anli 

^  ■   .    Tj-  —  tj- — -"-"3"?   JaLilo. 

Ex.  2.  Divide  j  by  ^. 

5    .      2    5V15 7  5 A3       A„Q 

Ex.  3.  Divide  -^  by  f .  ^rcs.  ^. 

Ex.  4.  Divide  §  by  2.  <£*&  ^ 

Ex.  5.  Divide  f  of  J  by  $  of  7-f.  jitaft  yf,-. 


RULE  OF  THREE  IN  VULGAR  FRACTIONS. 

Rule. — Prepare  the  fractions,  when  necessary,  as  previous- 
ly required ;  invert  the  first  term,  and  multiply  it  and  the  sec- 
ond and  third  terms  continually  together ;  the  product  will  be 
the  result  required. 

Example. — If  f  of  a  barrel  cost  -§  of  a  dollar,  what  will 
-^  of  a  barrel  cost  ? 


f:f::-rV—    ^X^X^=i=$0  33+,  Ans. 


2 

2 
Ex.  2.  What  will  3f  ounces  of  silver  cost  at  -^  of  a  pound 
sterling  per  ounce? 

Q  3 27  19   «f  20 380 38 19 

9 

-  X  -y-  X  —  =: shillings. 

,8       8        8 

fci|ixi^-2^i:=256||)«,or£l   Is.  ±\d.,Ans. 

Ex.  3.  What  part  of  a  ship  is  worth  $60,120,  when  J  of 

her  cost  $17,535 1  flfo+itf-^  Am. 


28  DECIMALS. 


DECIMALS. 

A  Decimal  is  a  fraction  which  has  for  its  denominator  a 
unit  with  as  many  ciphers  annexed  as  the  numerator  has 
places ;  it  is  usually  expressed  by  setting  down  the  numerator 
only,  with  a  point  on  the  left  of  it.  Thus,  -^  is  .4,  ■££$  is  .85, 
tottdit  is  -OO^,  and  towW  is  -00125.  When  there  is  a  de- 
ficiency of  figures  in  the  numerator,  ciphers  are  prefixed  to 
make  up  as  many  places  as  there  are  ciphers  in  the  denomi- 
nator. 

Mixed  Numbers  consist  of  a  whole  number  and  a  fraction, 
as,  3.25,  which  is  the  same  as  o.y2^,  or  -f-§-§. 

Ciphers  on  the  right  hand  make  no  alteration  in  the  value 
of  a  decimal,  for  .4,  .40,  .400  are  all  of  the  same  value,  each 
being  equal  to  -^j-.  • 

ADDITION  OF  DECIMALS. 

Rule. — Set  the  numbers  under  each  other,  according  to 
the  value  of  their  places,  as  in  whole  numbers,  in  which  po- 
sition the  decimal  points  will  stand  directly  under  each  other. 
Then,  beginning  at  the  right  hand,  add  up  all  the  columns  as 
in  whole  numbers,  and  place  the  point  directly  below  all  the 
other  points. 

Example.  —  Add  together  25.125,  56.19,  1.875,  and 
293.7325. 

25.125 
56.19 
1.875 
293.7325 
376.9225  the  Sum. 
Ex.  2.  Add  together  27.62,  .358,  17.3,  .007,  and  173.1. 

Sum  218.385. 
Ex.  3.  Add  together  .001,  .09,  and  .909.  Sum  1.000. 
Ex.  4.  Add  together  87.5,  56.25,  37.5,  and  43.75. 

Sum  225. 


DECIMALS.  29 


SUBTEACTION  OF  DECIMALS. 

Rule. — Set  the  numbers  under  each  other,  as  in  addition ; 
then  subtract  as  in  whole  numbers,  and  point  off  the  decimals 
as  in  the  last  rule. 

Example.— Subtract  15.150  from  89.1759. 
89.1759- 
15.150 

74.0259,  the  Rem. 
Ex.  2.  Subtract  96.50  from  100.  Rem.  3.50. 

Ex.  3.  Subtract  3.1416  from  4.5236.  Rem.  1.3820. 

Ex.  4.  Subtract  14.56789  from  2486.173. 

Rem.  2471.60511. 


MULTIPLICATION  OF  DECIMALS. 

Rule. — Set  the  factors,  and  multiply  them  together  the 
same  as  if  they  were  whole  numbers ;  then  point  off  in  the 
product  just  as  many  places  of  decimals  as  there  are  decimals 
in  both  the  factors.  But  if  there  are  not  so  many  figures  in 
the  product  as  there  are  decimal  places  required,  supply  the 
deficiency  by  prefixing  ciphers. 
Example. — Multiply  1.56  by  .75. 

1.56  , 

.75 
780 
1092 

1.1700,  the  Product. 
Ex.  2.  Multiply  79.25  by  .460.  Product  36.455. 

Ex.  3.  Multiply  79.347  by  23.15. 

Product  1836.88305. 
Ex.  4.  Multiply  .385746  by  .00464. 

Product  .00178986144. 


30  DECIMALS. 


BY   CONTRACTION. 


To  contract  the  Operation  so  as  to  retain  only  as  many  Decimal 
places  in  the  Product  as  may  be  thought  necessary. 

Rule. — Set  the  unit's  place  of  the  multiplier  under  the 
figure  of  the  multiplicand  whose  place  is  the  same  as  is  to  be 
retained  for  the  last  in  the  product,  and  dispose  of  the  rest  of 
the  figures  in  the  contrary  order  to  what  they  are  usually 
placed  in.  Then,  in  multiplying,  reject  all  the  figures  that 
are  more  to  the  right  hand  than  each  multiplying  figure,  and 
set  down  the  products,  so  that  their  right-hand  figures  may 
fall  in  a  column  directly  below  each  other;  and  observe  to 
increase  the  first  figure  in  every  line  with  what  would  have 
arisen  from  the  figures  omitted;  thus,  add  1  for  every  result 
from  5  to  14,  2  from  15  to  24,  3  from  25  to  34,  4  from  35 
to  44,  &c,  &c.,  and  the  sum  of. all  the  lines  will  be  the  prod- 
uct as  required. 

Example.— Multiply  13.57493  by  46.20517,  and  retain 
only  four  places  of  decimals  in  the  product. 

13.574  93 

71  502.64 


54  299  72 

8  144  96  +  2  for  18 

27150+2 

"    18 

6  79+4 

"   35 

14+1 

"     5 

9  +  2 

"   21 

627.23  20 

6  is  the  unit  of  the  multiplier,  and  9  is  the  figure  of  the  multi- 
plicand whose  place  is  the  same  as  is  to  be  retained  for  the  last  in 
the  product 

Ex.  2.  Multiply  27.14986  by  92.41035,  and  retain  only 
five  places  of  decimals.  Product  2508.92806. 

Ex.  3.  Multiply  480.14936  by  2.72416,  and  retain  only 
four  places  of  decimals.  Product  1308.0035. 


DECIMALS.  31 

Ex.  4.  Multiply  325.701428  by  .7218393,  and  retain  only 
three  places  of  decimals.  Product  235.103. 

Ex.  5.  Multiply  81.4632  by  7.24G51,  retaining  only  three 
places  of  decmials.  Product  590.324. 

DIVISION  OF  DECIMALS. 

Rule. — Divide  as  in  whole  numbers,  and  point  off  in  the 
quotient  as  many  places  for  decimals  as  the  decimal  places  in 
the  dividend  exceed  those  in  the  divisor ;  but  if  there  are  not 
so  many  places,  supply  the  deficiency  by  prefixing  ciphers. 
Example. — Divide  53.00  by  6.75. 
6.75)53.00(7.85  + 
47  25 
5  750 
5  400 


3500 
3375 


125 
Here  2  ciphers  were  annexed  to  carry  out  the  division. 
Ex.  2.  Divide  45.5  by  2100.  Quotient  .0216  +  . 

Ex.  3.  Divide  12  by  .7854.  Quotient  15.278. 

Ex.  4.  Divide  .061  by  79000. 

Quotient  .00000077215+. 
Ex.  5.  Divide  2.7182818  by  3.1415927. 

fc  Quotient  .865256—. 
Ex.  6.  Divide  .00128  by  8.192.  Quotient  .000156. 

BY   CONTRACTION. 

Rule. — Take  only  as  many  figures  of  the  divisor  as  will  be 
equal  to  the  number  of  figures,  both  integers  and  decimals,  to 
be  in  the  quotient,  and  find  how  many  times  they  may  be  con- 
tained in  the  first  figures  of  the  dividend,  as  usual. 

Let  each  remainder  be  a  new  dividend ;  and  for  every  such 
dividend  leave  out  one  figure  more  on  the  right-hand  side  of 
the  divisor,  carrying  for  the  figures  cut  off  as  in  Contraction 
of  Multiplication. 


32  DECIMALS. 

Note. —  When  there  are  not  so  many  figures  in  the  divisor  as  are  required 
to  be  in  the  quotient,  continue  the  first  operation  until  the  number  of  figures  in 
the  divisor  be  equal  to  those  remaining  to  be  found  in  the  quotient,  after  which 
begin  the  contraction. 

Example.— Divide  2508.92806  by  92.41035,  retaining  only 
four  places  of  decimals  in  the  quotient. 

92.4103|5)2508.928|06(27.1498 
1848  207+1 
660  721 
646  872+2 
13  849 
9  241 


4  608 
3  696 


912 

832+4 
80 

74+2 
6 
Ex.  2.  Divide  4109.2351  by  230.409,  retaining  only  four 
decimals  in  the  quotient.  Quotient  17.8345. 

Ex.  3.  Divide  37.10438  by  5713.96,  retaining  only  five 
decimals  in  the  quotient.  Quotient  .00649. 

Ex.  4.  Divide  913.08  by  2137.2,  retaining  only  three  deci- 
mals in  the  quotient.  Quotient  .427. 

REDUCTION  OF  DECIMALS. 

To  reduce  a  Vulgar  Fraction  to  its  equivalent  Decimal. 

Rule. — Divide  the  numerator  by  the  denominator,  as  in 
division  of  decimals,  annexing  ciphers  to  the  numerator  as 
far  as  necessary,  and  the  quotient  will  be  the  decimal  re- 
quired. 

.  Example. — Reduce  \  to  a  decimal. 
5)^0 

.8,  Quotient. 


DECIMALS.  33 

Ex.  2.  Keduce  -^j-  to  a  decimal. 

700)35.00(.05,  Quotient 
35  00 
Ex.  3.  Reduce  4  to  a  decimal.  Quotient  .625. 

Ex.  4.  Reduce  44  to  a  decimal.  Quotient  .9375. 

Ex.  5.  Reduce  -j-f-g-  to  a  decimal.  Quotient  .03125. 

To  find  the  Value  of  a  Decimal  in  Terms  of  an  in- 
ferior Denomination. 

Rule. — Multiply  the  decimal  by  the  number  of  parts  in  the 
next  lower  denomination,  and  cut  off  as  many  places  for  a  re- 
mainder to  the  right  hand  as  there  are  places  in  the  given 
decimal. 

Multiply  that  remainder  by  the  parts  in  the  next  lower 
denomination,  again  cutting  off  for  a  remainder,  and  so  on 
through  all  the  denominations  of  the  decimal. 

Then  the  several  denominations  pointed  off  on  the  left  hand 
will  give  the  result  required. 

Example. — What  is  the  value  of  .875  dollar? 
.875 
100 


87.500  cents, 

10 

5.000  milk. 

Ans.  87  cents,  5  mills. 

Ex.2. 

What  is  the  content  of  .140  cubic  foot  in  inches? 

.140 

1728  cubic  inches  in  a  cubic  foot 

241.920 

Ans.  241.920  cubic  inches. 

Ex.3. 

What  is  the  value  of  .00129  of  a  foot? 

Ans.  .01548  inches. 

Ex.4. 

What  is  the  value  of  1.075  ton  in  pounds? 

Ans.  2408. 

Ex.5. 

Reduce  .0125  lb.  troy  to  pennyweights. 

Ans.  3  dwts. 

B2 

34  DECIMALS. 

Ex.  6.  Reduce  .95  mile  to  its  equivalent  decimals  in  its 
lower  denominations. 
.95 
8  furlongs, 
TM 

40  rods. 
24.00  Ans.  7  furlongs  and  24  rods. 

Ex.  7.  Reduce  .05  mile  to  its  equivalent  decimals. 

Ans.  0  furlongs  and  16  rods. 
Ex.  8.  Reduce  ^  of  a  mile  to  its  equivalent  decimals. 

Ans.  §  furlongs  and  16  rods. 
Ex.  9.  Reduce  ^  of  a  cubic  yard  to  its  equivalent  decimals. 

Ans.  2.9999  feet+. 

Ex.  10.  Reduce  J  of  a  degree  to  its  equivalent  decimals. 

Ans.  19  minutes  and  59.999  seconds -f. 


To  reduce  a  Decimal  to  its  equivalent  in  a  higher  De- 
nomination. 

Rule. — Divide  by  the  number  of  parts  in  the  next  higher 
denomination,  continuing  the  operation  as  far  as  required. 
Example. — Reduce  1  inch  to  the  decimal  of  a  foot. 


12 


1.00000 


.08333,  &c,  Ans. 
Ex.  2.  Reduce  14  minutes  to  the  decimal  of  a  day. 


60 
24 


14.00000 


.23333 


.00972,  &c,  Ans. 
Ex.  3.  Reduce  14"  12'"  to  the  decimal  of  a  minute. 
14"  12/7/ 
60 


60 
60 


852.' 


X4.2y 


.23666',  &c,  Ans. 


Note. — When  there  are  several  numbers,  to  be  reduced  all  to  the  decimal 
of  the  highest. 


DECIMALS.  35 

Reduce  them  all  to  the  lowest  denomination,  and  proceed 
as  for  one  denomination. 

Ex.  4.  Reduce  5  feet  10  inches  and  3  barleycorns  to  the 
decimal  of  a  yard. 

Feet.  Inches.    Be. 

5     10    3 
12 
70 

3 


213. 


71. 


5.9166 


1.9722,  &c,  Ans. 
Ex.  5.  Reduce  1  dwt.  to  the  decimal  of  a  pound  troy. 

Ans.  .004166+  lb. 
Ex.  6.  Reduce  1  yard  to  the  decimal  of  a  mile. 

Ans.  .000568+  mile. 
Ex.  7.  Reduce  8  feet  6  inches  to  the  decimal  of  a  mile. 

Ans.  .0016098  mile. 
Ex.  8.  Reduce  4J  miles  to  the  decimal  of  80  miles. 

Ans.  .05625. 
Ex.  9.  Reduce  14',  18",  and  36'"  to  the  decimal  of  a  de- 
gree.    •  Ans.  .2385  degree. 

Ex.  10.  Reduce  17  yards,  1  foot,  and  5.98848  inches  to  the 
decimal  of  a  mile.  Ans.  .009943  mile. 


RULE  OF  THREE  IN  DECIMALS. 

Rule. — Prepare  the  terms  by  reducing  vulgar  fractions  to 
decimals,  compound  numbers  to  decimals  of  the  highest  de- 
nominations, and  the  first  and  third  terms  to  the  same  denom- 
ination ;  then  proceed  as  in  whole  numbers. 

Example. — If  ^  a  ton  of  iron  cost  f  of  a  dollar,  what  will 
.625  of  a  ton  cost? 

i  =  .5  \  .5:. 75::. 625 

:=.75j 


|=.75j  .625 

.5).46875 


.9375,  Ans. 


36  DUODECIMALS. 

Ex.  2.  If  -§  of  a  yard  cost  §  of  a  dollar,  what  will  ^  of  a 
yard  cost?  Am.  .3333+  dollar. 

Ex.  3.  If  t^t  of  a  mile  cost  $15.75,  what  will  ^  of  a  fur- 
long cost?  Ans.  $4  05. 


DUODECIMALS. 


In  Duodecimals,  or  Cross  Multiplication,  the  dimensions  are 
taken  in  feet,  inches,  and  twelfths  of  an  inch. 

Rule. — Set  down  the  dimensions  to  be  multiplied  together, 
one  under  the  other,  so  that  feet  may  stand  under  feet,  inches 
under  inches,  &c. 

Multiply  each  term  of  the  multiplicand,  beginning  at  the 
lowest,  by  the  feet  in  the  multiplier,  and  set  the  result  of  each 
directly  under  its  corresponding  term,  carrying  1  for  every  12 
from  1  term  to  the  other. 

In  like  manner,  multiply  all  the  multiplicand  by  the  inches 
of  the  multiplier,  and  then  by  the  twelfth  parts,  setting  the 
result  of  each  term  one  place  farther  to  the  right  hand  for 
every  multiplier.  The  sum  of  the  products  is  the  result  re- 
quired. 

Example. — Multiply  1  foot  3  inches  by  1  foot  1  inch. 

Feet.    Inches. 

1      3 
1      1 


3 


14      3 
Proof. — 1  foot  3  inches  is  15  inches,  1  foot  1  inch  is  13 
inches;    and  15x13  =  195   square  inches.     Now  the  above 
product  reads  1  foot,  4  inches,  and  3  twelfths  of  an  inch,  and 
1  foot        =144  square  inches. 
4  inches    =48  " 

3  twelfths  =     3  " 

195  " 

which  is  the  product  required. 


DUODECIMALS.  37 

Ex.  2.  How  many  square  feet,  inches,  &c,  are  there  in  a 
platform  35  feet  4^  inches  long,  and  12  feet  3 J  inches  wide? 


Feet. 

35 

Inches. 

4 

Twelfths. 

6 

12 

3 

4 

424 

6 

8 

10 

1 

6 

11 

9 

6 

0 

434      3       11         0         0. 

Or  434  feet,  3  inches,  and  11  twelfths. 
Ex.  3.  Multiply  20  feet  6£  inches  by  40  feet  6  inches. 
Ans.  831  feet,  11  mcAes,  3  twelfths,  which  is  equal  to  831 
square  feet  and  135  square  inches. 

By  decimals,  40  ft.  6    in. =40.5 

20  ft.  6  J  in.  =  20541666,  &c. 

831.937499  square  feet. 
144 


134.999856  square  inches. 

Table  showing  the  value  of  Duodecimals  in  Square 
Feet  and  Decimals  of  an  Inch. 


1  Foot 

1  Inch 

1  Twelfth 
^  of  1  Twelfth 
-jL  of  -j^  of  1  Twelfth 

Sq.  Feet.                     Sq.  Inches. 

1      or  144. 

.  1          «          1 

i        «           08W 

17  2tJ                         .VOOOc 

1        "            OOfi<V 

20  73  6                   .UUOJ^ 

Application  of  this  Table. 

What  number  of  square  inches  are  there  in  a  floor  100J 
feet  broad,  and  25  feet,  6  inches,  and  6  twelfths  long  % 
100^  feet  =   100.5  /jig. 

~25      «  (25xl44)=3600.    iwctes. 

6kk  (6x12)     =     72.       " 

6  fegeffifo  =; 6.       " 

25   feet  6  wcfos  6  twelfths  =3678.    inches. 


AOPTHE  "^> 


38  INVOLUTION. 

12)3678.  inches. 
12)  306.5 


2544166  feet* 

As  the  3678  are  square  inches,  it  is  necessary  to  divide  by  144  to  produce 
square  feet,  and  the  operation  is  more  readily  performed  by  dividing  twice  by 
12  (12  X  12  =  144)  than  by  144  in  one  division. 

Then  25.54166 

100.5 
t  feet  2566.936830 

12 


inches  11.241960 
12 


twelfths  2.903520 

Or, 2566 /art. 

11x12      =  132.      inches. 

2  =  2.         " 

.9  (.9-^12)= .75     " 

2566  feet,  134.75  inches. 


INVOLUTION. 

Involution  is  the  multiplying  any  number  into  itself  a  cer- 
tain number  of  times.  The  products  obtained  are  called 
Powers.     The  number  is  called  the  Root,  or  first  power. 

When  a  number  is  multiplied  by  itself  once,  the  product  is 
the  square  of  that  number ;  twice,  the  cube ;  three  times,  the 
biquadrate,  &c.     Thus,  of  the  number  5, 

5  is  the  Root,  or  1st  power. 
5x5=25     "      Square,  or   2d   power,  and   is   ex- 
pressed 52. 
5x5x5  =  125     "      Cube,  or  3d  power,  and  is  expressed  53. 
5x5x5x5  =  625     "      Biquadrate,  or  4th  power,  and  is  ex- 
pressed 54. 
The  little  figure  denotes  the  power,  and  is  called  the  Index 
or  Exponent.  - 


EVOLUTION.  39 

Example. — What  is  the  cube  of  9  ? 

9x9x9=729,^715. 
Ex.  2.  What  is  the  cube  of  1 1  Ans.  |£. 

Ex.  3.  What  is  the  4th  power  of  1.5  ?  Ans.  5.0625. 

Ex.  4.  What  is  the  square  of  4.16  ?  Am.  17.3056. 

Ex.  5.  What  is  the  square  off?  Ans.  J. 

Ex.  6.  What  is  the  third  power  of  f 1  Ans.  fjf . 

Ex.  7.  What  is  the  fourth  power  of  .025  ! 

Ans.  .000000390625. 
Ex.  8.  What  is  the  fifth  power  of  5  !  Ans.  3125. 

Ex.  9.  What  is  the  fifth  power  of  .05  ! 

Ans.  .0000003125. 


EVOLUTION. 

Evolution  is  finding  the  Root  of  any  number. 
The  sign  -y/  placed  before  any  number  indicates  the  square 
root  of  that  number  is  required  or  shown. 

The  same  character  expresses  any  other  root  by  placing  the 
index  above  it. 

Thus,  <t/25=5,         4+2  =  ^36. 
^27=3,  and  {/6±  =       4. 
Roots  which  only  approximate  are  called  Surd  Roots,  but 
those  which  are  exact  are  called  Rational  Roots. 

TO  EXTRACT  THE  SQUARE  ROOT. 

Rule. — Point  off  the  given  number  from  the  place  of  units 
into  periods  of  two  figures  each  by  setting  a  point  over  the 
place  of  units,  another  over  that  of  hundreds,  and  so  on  over 
every  second  figure,  both  to  the  left  hand  in  integers  and  to 
the  right  hand  in  decimals. 

Find  the  greatest  square  in  the  left-hand  period,  and  place 
its  root  in  the  quotient ;  subtract  the  square  number  from  the 
left-hand  period,  and  to  the  remainder  bring  down  the  next 
period  for  a  new  dividend. 

Double  the  root  already  found  for  a  divisor ;  find  how  many 


40 


EVOLUTION. 


times  this  incomplete  divisor  is  contained  in  the  dividend  ex- 
clusive of  the  right-hand  figure ;  with  the  consideration  that 
the  result,  or  root,  is  to  be  the  units'  figure  of  the  complete  di- 
visor, place  the  result  in  the  quotient,  and  at  the  right  hand 
of  the  divisor. 

Multiply  this  divisor  by  the  last  quotient  figure,  and  sub- 
tract the  product  from  the  dividend;  bring  down  the  next 
period,  and  proceed  as  before. 

Example. — What  is  the  square  root  of  2  ? 


2.000000(1.414,  &c,  Ans. 

1 


24 
4 


100 
96 


281 

1 


400 

281 


2824 
4 


2828 


11900 
11296 


604 


Ex.  2.  What  is  the  square  root  of  144  ? 


144(12,  Ans. 
1 


22 


044 
44 


Ex.  3.  What  is  the  square  root  of  17.3056  ! 


17.3056(4.16,  Ans. 
16 


130 
81 


826 


4956 
4956 


Ex.  4.  What  is  the  square  root  of  .000729  1     Ans.  .027. 


SQUARE   ROOTS    OF   VULGAR   FRACTIONS. 

Rule. — Reduce  the  fractions  to  their  lowest  terms,  and 
that  fraction  to  a  decimal,  then  proceed  as  in  whole  numbers 
and  decimals. 


EVOLUTION.  41 

Note. —  When  the  terms  of  the  fractions  are  squares,  take  the  root  of  each 
and  set  one  above  the  other ;  as,  the  square  root  of^  is  %. 

Example. — What  is  the  square  root  of  ^^ 
-^=.41666666,  &c 
6    .4i666666(.6454,  Ans. 
6     36 


124 

4 


566 
496 


1285 
5 
12904 

4 


7066 

6425 
64166 
51616 


12908   12550 
Ex.  2.  What  is  the  square  root  of  -^  I 

Ans.  0.8660254. 
Ex.  3.  What  is  the  square  root  of  17-f  ? 

171=17x84-3=-^^  and  i§£=17.875, 
which  is  the  integer  and  decimal  from  which  the  root  is  to  be 
extracted.  .  Ans.  4.1683,  &c. 


TO   EXTRACT   THE   CUBE   ROOT 

By  the  ordinary  Method. 

Point  off  the  given  number  from  the  place  of  units  into 
periods  of  three  figures  each,  by  setting  a  point  over  the  place 
of  units,  and  another  also  over  every  third  figure  from  thence 
to  the  left  hand  in  integers,  and  to  the  right  hand  in  decimals. 

Find  the  greatest  cube  in  the  left-hand  period,  and  set  its 
root  in  the  quotient ;  subtract  the  cube  number  from  the  left- 
hand  period,  and  to  the  remainder  bring  down  the  second  pe- 
riod for*a  new  dividend. 

To  three  times  the  square  of  the  root  just  found  add  three 
times  the  root  itself,  setting  this  one  figure  place  more  to  the 
right  than  the  former ;  add  these  products  together,  and  call 
their  sum  the  divisor.  Divide  the  new  dividend,  less  the  last 
figure  to  the  right  hand,  by  the  divisor  for  the  next  figure  of 


42 


EVOLUTION. 


the  root,  which  annex  to  the  former,  calling  this  last  figure  e, 
and  the  part  of  the  root  before  found  a. 

Add  the  following  three  products  together,  viz.,  thrice  a 
square  multiplied  by  e,  thrice  a  multiplied  by  e  square,  and  e 
cube,  setting  each  of  them  one  figure  place  more  to  the  right 
than  the  former,  and  call  their  sum  the  subtrahend;  which 
must  not  exceed  the  dividend,  but  if  it  does,  then  make  the 
last  figure  e  less,  and  repeat  the  operation  for  finding  the  sub- 
trahend till  it  be  less  than  the  dividend.         • 

From  the  dividend  take  the  subtrahend,  and  to  the  remain- 
der join  the  next  period  of  the  given  number  for  a  new  divi- 
dend, to  which  form  a  new  divisor  from  the  whole  root  already 
formed,  and  from  thence  another  figure  of  the  root  as  already 
directed,  and  so  on  until  the  extraction  is  complete. 

Example. — Extract  the  cube  root  of  48228.544. 


3x32: 
3x3  : 

divisor 


:27 
:     09 


279 

3x32x6  =162     " 
3x3  x62=  324 
63  =     216, 
3x362r=3888 
3x36  =     108 
divisor 
3X362x4 
3x36 


48228.544(36.4,  root. 
21228  dividend. 


19656  subtrahend. 


38988 


15552 
X42=  1728 
43=     64 


1572544  dividend. 


1572544  subtrahend. 


000  remainder. 
Ex.  2.  Extract  the  cube  root  of  46656.  Ans.  36. 

Ex.  3.  Extract  the  cube  root  of  46656000.       Ans.  360. 
Ex.  4.  Extract  the  cube  root  of  2.  Ans.  1.259921. 

Ex.  5.  Extract  the  cube  root  of -j-^.  Ans.  .25. 


TO  EXTRACT  THE  CUBE  ROOT. 

RUle. — From  a  table  of  Roots  take  the  nearest  cube  to  the 
given  number,  and  call  it  the  assumed  cube. 


EVOLUTION.  43 

Then,  as  the  given  number  added  to  twice  the  assumed  cube 
is  to  the  assumed  cube  added  to  twice  the  given  number,  so  is 
the  root  of  the  assumed  cube  to  the  required  cube  root,  nearly. 
By  using,  in  like  manner,  the  root  thus  found  as  an  as- 
sumed cube,  and  proceeding  as  above,  another  root  will  be 
found  still  nearer,  and  in  the  same  manner  as  far  as  may  be 
necessary. 

Example. — What  is  the  cube  root  of  10517.9  ? 
Nearest  cube  10648,  root  22. 
10648.       10517.9 

2_     2 

21296    21035.8 

10517.9  10648. 

31813.9  :  31683.8::  22  :  21.9+,  Ans. 

Tafind  the  fourth  Root  of  a  Number. 

Rule. — Extract  the  square  root  twice. 

Example.— What  is  the  4th  root  of  625  ?  Ans.  5. 

To  find  the  sixth  Root  of  a  Number. 

Rule. — Take  the  cube  root  of  its  square  root. 
Example.— What  is  the  ■{/  of  441? 

V441  =  21,  and^21  =  2.758923,  Ans. 

To  find  the  eighth  Root  of  a  Number. 

Rule. — Extract  the  square  root  thrice. 

Example.— What  is  the  8th  root  of  390625 1       Ans.  5. 

To  extract  any  Root  whatever. 
Let  P  represent  the  number, 

n  "        the  index  of  the  power, 

A  '       "        the  assumed  power,  r  its  root. 
R         "        the  required  root  of  P. 
Then,  as  the  sum  of  n-f-1  x  A  and  n— 1  xP  is  to  the  sum 
of  n-f-1  xP  and  n— 1  x  A,  so  is  the  assumed  root  r  to  the 
required  root  R. 


44  PROPERTIES    OP  NUMBERS. 

Example. — What  is  the  cube  root  of  1500  ? 
The  nearest  cube  is  1331,  root  11. 

P=1500,  n=3,  A=1331,  r=ll; 
then, 

«+lxA=5324,  7i+lxP=6000 
7z-lxF-3000,  yz-lxA-2662 

8324  :  8662  ::  11 :  11.446+,  Ans. 


PEOPERTIES  OF  NUMBERS. 

1.  A  Prime  Number  is  that  which  can  only  be  measured 
(divided  without  a  remainder)  by  1  or  unity. 

2.  A  Composite  Number  is  that  which  can  be  measured  by 
some  number  greater  than  unity. 

3.  A  Perfect  Number  is  that  which  is  equal  to  the  sum  of 
all  its  divisors  or  aliquot  parts;  as  6=-§,  •§-,  -§. 

4.  If  the  sum  of  the  digits  constituting  any  number  be  di- 
visible by  3  or  9,  the  whole  is  divisible  by  them. 

5.  A  square  number  can  not  terminate  with  an  odd  num- 
ber of  ciphers. 

6.  No  square  number  can  terminate  with  two  equal  digits, 
except  two  ciphers  or  two  fours. 

7.  No  number  the  last  digit  of  which  is  2,  3,  7,  or  8,  is  a 
square  number. 


GEOMETRY.  45 


GEOMETRY. 

Definition^, 

For  the  Definitions  of  the  Surfaces  of  Figures,  Solids,  Lines, 
&c,  &c,  see  Mensuration  of  Areas,  Lines,  and  Surfaces,  Men- 
suration of  Solids  and  Conic  Sections. 

A  Point  has  position,  but  not  magnitude. 

A  Line  is  length  without  breadth,  and  is  either  Right,  Curved,  or 
Mixed.  \ 

A  Right  Line  is  the  shortest  distance  between  two  points. 

A  Mixed  Line  is  composed  of  a  right  and  a  curved  line. 

A  Superficies  has  length  and  breadth  only,  and  is  plane  or  curved. 

A  Solid  has  length,  breadth,  and  thickness. 

An  Angle  is  the  opening  of  two  lines  having  different  directions,  and 
is  either  Right,  Acute,  or  Obtuse. 

A  Right  Angle  is  made  by  a  line  perpendicular  to  another  falling 
upon  it. 

An  Acute  Angle  is  less  than  a  right  angle. 

An  Obtuse  Angk  is  greater  than  a  right  angle. 

An  Arc  is  any  part  of  the  circumference  of  a  circle. 

A  Chord  is  a  right  line  joining  the  extremities  of  an  arc. 

The  Radius  of  a  circle  is  a  line  drawn  from  the  centre  to  the  circum- 
ference. 

A  Semicircle  is  half  a  circle. 

A  Quadrant  is  a  quarter  of  a  circle. 

A  Secant  is  a  line  that  cuts  a  circle,  lying  partly  within  and  partly 
without  it. 

A  Cosecant  is  the  secant  of  the  complement  of  an  arc. 

A  Sine  of  an  arc  is  a  line  running  from  one  extremity  of  an  arc  per- 
pendicular to  a  diameter  passing  through  the  other  extremity,  and  the 
sine  of  an  angle  is  the  sine  of  the  arc  that  measures  that  angle.   , 

The  Versed  Sine  of  an  arc  or  angle  is  the  part  of  the  diameter  inter- 
cepted between  the  sine  and  the  arc. 

The  Cosine  of  an  arc  or  angle  is  the  part  of  the  diameter  intercepted 
between  the  sine  and  the  centre. 

A  Tangent  is  a  right  line  that  touches  a  circle  without  cutting  it. 

A  Cotangent  is  the  tangent  of  the  complement  of  the  arc. 

The  Circumference  of  every  circle  is  supposed  to  be  divided  into  360 
equal  parts,  called  Degrees ;  each  degree  into  60  Minutes,  and  each  min- 
ute into  60  Seconds,  and  so  on. 

The  Complement  of  an  angle  is  what  remains  after  subtracting  the 
angle  from  90  degrees. 

The  Supplement  of  an  angle  is  what  remains  after  subtracting  the 
angle  from  180  degrees. 

Note. — A  Triangle  is  also  called  a  Trigon,  and  a  Square  a  Tetragon. 


46 


GEOMETRY. 


To  exemplify  these  definitions,  let  A  c  b,  in  the  following  diagram, 
be  an  assumed  arc  of  a  circle  described  with  the  radius  A  B. 


B  k,  the  Cosine  of  the  arc  A  c  b. 

A  g,  the  Tangent  of  do. 

C  B  6,  the  Complement,  and  b  B  E, 
the  Supplement  of  the  arc  Ac  b. 

C  g,  the  Cotangent  of  the  arc,  writ- 
ten cot. 

B  g,  the  Cosecant  of  the  arc,  written 
cosec. 

m  C,  the  Cover sed  sine  of  the  arc, 
or,  by  convention,  of  the  angle 
A  B  b :  written  coversin. 


A  c  &,  an  Arc  of  the  circle  A  C  E  D. 

A  6,  the  Chord  of  that  arc. 

e  D  a,  a  Segment  of  the  circle. 

A  B,  the  Radius. 

A  B  b,  a  Sector. 

A  D  E  B,  a  Semicircle. 

C  B  E,  a  Quadrant. 

A  e  a  E,  a  Zone. 

n  o  h,  a  Lune. 

B  g,  the  Secant  of  the  arc  A  c  b. 

b  k,  the  Sine  of  do. 

A  k,  the  Versed  Sine  of       do. 

The  Vertex  of  a  figure  is  its  top  or  upper  point.  In  conic  sections  it 
is  the  point  through  which  the  generating  line  of  the  conical  surface 
always  passes. 

The  Altitude,  or  height  of  a  figure,  is  a  perpendicular  let  fall  from  its 
vertex  to  the  opposite  side,  called  the  base. 

The  Measure  of  an  angle  is  an  arc  of  a  circle  contained  between  the 
two  lines  that  form  the  angle,  and  is  estimated  by  the  number  of  de- 
gi-ees  in  the  arc.  '  • 

A  Segment  is  a  part  cut  off  by  a  plane,  parallel  to  the  base. 

A  Frustrum  is  the  part  remaining  after  the  segment  is  cut  off. 

The  Perimeter  of  a  figure  is  the  sum  of  all  its  sides. 

A  Problem  is  something  proposed  to  be  done. 

A  Postulate  is  something  required. 

A  Theorem  is  something  proposed  to  be  demonstrated. 

A  Lemma  is  something  premised,  to  render  what  follows  more  easy. 

A  Corollary  is  a  truth  consequent  upon  a  preceding  demonstration. 

A  Scholium  is  a  remark  upon  something  going  before  it. 


MENSURATION    OF   AREAS,  LINES,  AND    SURFACES. 


47 


MENSURATION  OF  AREAS,  LINES,  AND  SUR- 
FACES. 

OF    FOUR-SIDED    FIGURES. 

! 

Parallelograms. 
Definition.     Quadrilaterals  having  their  opposite  sides  parallel. 

To  ascertain  tlie  Area  of  a  Square,  a  Rectangle,  a  Rhombus,  or  a 
Rhomboid. 

Rule. — Multiply  the  length  by  the  breadth  or  height,  and 
the  product  will  be  the  area. 
Or,  length  x  breadth  — area. 

Note. — The  surface  of  any  Quadrilateral  is  equal  to  half  the  product 
of  the  diagonals  X  the  sine  of  their  angle. 

Or,  abxb  c=area. 

Or,  Ixb—area,  I  representing  the  length  and  b  the  breadth. 


Square. 
Definition.  A  plane  superficies  having  equal  sides  and  angles. 

Fig.  1.      a 
b 

X 

c 

Example. — The  sides  a  b,  b  c,fig.  1,  are  5  feet  6  inches 
(5.5) ;  what  is  the  area. 

5.5x5.5  =  30.25  square  feet. 
Centre  of  Gravity.  Is  in  its  geometrical  centre,*  o. 

*  The  geometrical  centre  of  any  Parallelogram,  regular  Polygon, 
Circle,  Ellipse,  &c,  &c,  is  the  point  of  intersection  of  two  or  more  of 
their  respective  diagonals,  radii,  or  diameters.- 


48 


MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 


The  side  of  a  Square  is  equal  to  the  square  root  of  its  area. 

Example. — What  is  the  side  of  a  square  when  the  area  of 
it  is  1024  feet?  Ans.  32  feet. 

Rectangle. 

Definition.     A  plane  Superficies  with  parallel  sides  and  equal 
angles. 

Fig.  2. 


Example. — The  side  a  &,  Jig.  2,  is  5  feet,  and  b  c  1  feet 
3  inches  (7.25);  what  is  the  area? 

5  x  7.25=36.25  square  feet. 

The  length  of  a  Rectangle  is  equal  to  the  area  divided  by  its  breadth. 

Example. — What  is  the  length  of  a  rectangle,  the  area 
being  2048  feet  and  the  breadth  32  !  Ans.  64  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre,  o. 

Rhombus  (Lozenge). 

Definition.  A  plane  superficies  with  equal  sides,  but  its  angles 
not  right  angles. 

Fig. 


Example. — The  height  a  b,  fig.  3,  is  5  feet  9  inches  (5.75), 
the  length  a  c,  7  feet ;  what  is  the  area  ? 

5.75x7=40.25  square  feet. 

Note. — The  opposite  angles  of  a  Rhombus  are  equal. 
Centre  of  Gravity.  Is  in  its  geometrical  centre,  o. 


MENSURATION    OF  AREAS,  LINES,  AND  SURFACES.  49 

Rhomboid. 

Definition.  A  plane  superficies  with  parallel  sides,  but  its  angles 
not  right  angles. 

Fig.  4.        a c 


b 
Example. — The  breadth  a  b,  Jig.  4,  is  5  feet,  and  the  length 
a  Cj  6  feet ;  what  is  the  area  1 

5x6=30  square  feet. 

The  opposite  angles  of  a  Rhomboid  are  equal. 
Centre  of  Gravity.  Is  in  its  geometrical  centre,  o. 

Gnomon. 

Definition.  The  space  included  between  the  lines  forming  two 
parallelograms,  of  which  the  smaller  is  inscribed  within  the  larger, 
so  that  the  angles  of  each  are  common  to  both. 

To  ascertain  the  area  of  a  Gnomon. 

.    Rule. — Find  the  areas  of  the  two  parallelograms,  and  sub- 
tract the  less  from  the  greater;  the  difference  will  give  the 
area  required. 
Or,  a—a/=:area. 

Example. — The  dimensions  of  a  gnomon  are  10  by  10  and 
6  by  6  inches ;  what  is  its  area  ? 

10x10=100.     6x6=36. 
Then,  100— 36  =  64 ^difference  of  areas  of  the  parallelograms 
—  the  area  required.  .     . 

Ex.  2.  The  dimensions  of  two  concentric  parallelograms 
are  15x8  and  12x6  feet ;  what  is  the  area  of  the  gnomon  ? 

Ans.  48  feet. 
Centre  of  Gravity.  Is  in  its  geometrical  centre. 
C 


50  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


Triangles. 
Definition.     Plane  superficies  having  three  sides  and  angles, 
and  are  designated 

Right  angled,  when  one  of  the  angles  is  a  right  angle. 
Acute  angled,  when  all  its  angles  are  less  than  a  right  angle. 
Obtuse  angled,  when  one  angle  is  greater  than  a  right  angle. 
Equilateral,  when  the  sides  are  equal. 
Isosceles,  when  two  of  the  sides  are  equal. 
Scalene,  when  all  the  sides  are  unequal. 

Notes. — Equiangular  triangles  are  similar,  or  have  their  like  sides 
proportional. 

The  Hypothenuse  is  that  side  of  a  right-angled  triangle  which  is  op- 
posite to  the  right  angle. 

The  perpendicular  height  of  a  triangle  is  equal  to  twice  its  area  di- 
vided hy  its  base. 

To  ascertain  the  Area  of  a  Triangle,  Figs.  5,  6,  7. 

Rule. — Multiply  the  base  a  b,  by  the  height  c  d,  and  half 
the  product  is  the  area. 

„     a bxc d 

Or, = = area. 


area,  b  representing  the  base,  and  h  the  height. 
c      Fig.  6.  c      Fi9-  7- 


)         a  d  bad  b 

Example. — The  base  a  b,fig.  5,  is  4  feet,  and  the  height 
c  b,  6  feet ;  what  is  the  area  % 

4x6=24,  and  24^-2  =  12  square  feet. 
Ex.  2.  The  base  a  b,  fig.  6,  is  5  feet  6  inches,  and  the  height 
c  d,  5  feet ;  what  is  the  area  ? 

5  feet  6  inches =5.5. 
5.5  x5=27.5,  and  27.5-^2  =  13.75  square  feet. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  51 

V 

Ex.  3.  The  base  a  b,  fig.  7,  is  6  feet  3  inches,  and  the  height 
c  d  is  5  feet  8  inches ;  what  is  the  area  ? 

6  feet  3  inches=6.25,  and  5  feet  8  inches=5.666. 

Then,  6.25  x  5.666=35.413,  and  35^13  =  17.706  square  feet. 

Ex.  4.  The  base  a  b,  fig.  7,  is  18  feet  4  inches,  and  the 

height  c  d,  10  feet  11  inches;  what  is  the  area? 

18  ft.  4  in.  =  18.333,  and  10  ft.  11  in.  =  10.916. 

200  12S 
18.333  x  10.916  =  200.123,  and  — ^ =  100.0625  feet. 

Ex.  5.  What  is  the  area  of  a  triangle  when  the  base  of  it  is 
20  feet  and  the  height  10.25  ?  Ans.  102.5  feet. 

Ex.  6.  When  the  height  of  a  triangle  is  16.75  feet  and  the 
base  6.24,  what  is  its  area?  Ans.  52.26  feet. 

Ex.  7.  What  is  the  area  of  a  right-angled  triangle  when  its 
two  shortest  sides  are  26  and  28  feet?  Ans.  364  feet. 

Ex.  8.  The  base  of  an  equilateral  triangle  is  10  feet,  what 
is  its  area  ? 

10-7-2=5,  or  the  length  of  the  base  of  two  equal  right-angled 
triangles,  of  which  the  length  of  the  remaining  sides  of  the  triangle 
is  iiie  hypothenuse. 

Therefore,  102— 52  =  75,  and  y75  =  8.66,  the  height  of  each 
of  the  right-angled  triangles. 

Hence,  8.66x5=43.3  square  feet. 

Ex.  9.  The  equal  sides  of  an  isosceles  triangle  are  50  feet, 
and  its  base  28  ;  how  many  square  yards  does  it  contain  ? 

Ans.  74f  yards. 

To  ascertain  the  area  of  a  Triangle  by  the  length  of  its  Sides 
{Figs.  6  and  7). 

Eule. — From  half  the  sum  of  the  three  sides  subtract  each 
side  separately ;  then  multiply  the  half  sum  and  the  three  re- 
mainders continually  together,  and  the  square  root  of  the  prod- 
uct is  the  area. 


Or,  Vs  (s—a)x{s—b)x(s—c)=area,  a,  b,  c  being  the  sides, 

.       a-{-b-{-c 
and  s= — . 


52  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

When  all  the  Sides  are  equal. 

Rule. — Square  the  length  of  a  side,  and  multiply  one  fourth 

of  the  product  by  1.732,  and  this  product  will  give  the  area 

required. 

S2 
Or,  —  x  1.732  =area. 

Example. — The  sides  of  a  triangle  are  30,  40,  and  50; 

what  is  the  area  in  square  feet  ? 

30+40  +  50     120     r(.        .   . .  ;,,      ., 

! ! = ==60,  or  half  sum  of  the  sides. 

2  2 

60—30=30^ 

60 — 40  =  20  y  remainders. 

60-50  =  10j 

Whence  30x20x10x60=360000,  and  ^360000=600 

square  feet. 

Ex.  2.  How  many  acres  are  there  in  a  triangle,  the  sides 

of  which  are  49,  50.25,  and  26  chains?  Ans.  62.1875. 

Ex.  3.  What  is  the  area  of  a  triangle,  the  three  sides  being 

26,  28,  and  30  feet?  Ans.  336  feet. 

Ex.  4.  What  is  the  area  of  a  triangle  when  its  sides  are 

each  50  feet? 

502 

— —  =z§2o  =one  fourth  of  the  square  of  the  length  of  a  side, 

and  625  x  1.732  =  1082.5  feet. 
Ex.  5.  What  is  the  area  of  a  right-angled  triangle  when 
its  sides  are  50,  50,  and  70.7107  feet?  Ans.  1250  feet. 

Ex.  6.  What  is  the  area   of  a  triangle  in  square  yards 
when  its  sides  are  500, 400,  and  300  feet? 

Ans.  6666.66  yards. 

To  ascertain  the  length  of  one  side  of  a  right-angled  Triangle,  the 
length  of  the  other  two  sides  being  given  (Fig  5). 

When  the  two  legs  are  given,  To  ascertain  the  hypothenuse. 

Rule. — Add  together  the  squares  of  the  two  legs  a  b,  b  c, 
and  extract  the  square  root  of  the  sum. 


MENSURATION    OP  AREAS,  LINES,  AND    SURFACES.  53 


Or,  V«  b2-\-b  c2  =  hypothenuse. 

Or,  Vb2  +  h2,  b  representing  the  base,  and  h  the  height. 
Example. — The  base  a  b  is  30  inches,  and  the  height  b  c 
40  ;  what  is  the  length  of  the  hypothenuse  % 

302+402  —  2500,  and  -/2500=50  inches. 
Ex.  2.  The  base  of  a  right-angled  triangle  is  38.5  feet,  and 
the  perpendicular  18  feet;  what  is  the  hypothenuse? 

Ans.  42.5  feet. 
Ex.  3.  The  height  of  a  church  steeple  is  103  feet,  and  the 
length  of  its  shadow  is  320  feet ;  what  is  the  distance  from 
the  point  of  the  shadow  to  the  top  of  the  steeple  ? 

Arts.  336.17  feet. 
Ex.  4.  The  base  of  a  triangle  is  14  feet,  and  the  height  of 
it  48  ;  what  is  the  length  of  its  hypothenuse  ? 

Ans.  50  feet. 
Ex.  5.  The  sides  of  a  triangle  are  6  yards,  1  foot,  and  11.4 
inches  ;  what  is  the  length  of  its  hypothenuse  ? 

Ans.  9  yards,  1  foot,  and  .2135  in. 

To  ascertain  the  other  Leg  {Fig.  5),  the  Hypothenuse  and  one  of 
the  Legs  being  given. 

Rule. — Subtract  the  square  of  the  given  leg  from  the  square 
of  the  hypothenuse,  and  the  square  root  of  the  remainder  is 
the  length  of  the  leg  required. 


Or,  ■jhyp.2-  -Lj^ 


Or    Ja  c2      Sab2  =  b  c- 
Or,  ya  c  -  |5  c2=a  h 

Example. — The  base  of  a  triangle  is  30  feet,  and  the  hy- 
pothenuse 50  ;  what  is  the  height  of  it? 

502-302z=2500-900,  and  2500-900  =  1000. 
Then  -/1600  =  40/ee*. 
Ex.  2.  The  hypothenuse  of  a  triangle  is  50  feet,  and  the 
perpendicular  40 ;  what  is  the  base  ? 

502-402  =  2500-1600,  and  2500-1600  =  900. 
Then  -/900= 30/^. 


54 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


Ex.  3.  It  is  required  to  find  the  length  of  a  ladder,  the 
lower  end  placed  15  feet  from  the  face  of  a  wall,  and  the  up- 
per end  resting  on  it  at  a  height  of  26  feet. 

Ans.  30.017  feet. 

Ex.  4.  A  pole*  50  feet  in  length,  being  placed  in  a  street, 
reached  the  sill  of  a  window  30  feet  from  the  ground  on  one  side, 
and  being  turned  over  without  removing  the  foot  from  its  lo- 
cation, it  reached  a  window  on  the  opposite  side  45^  feet  high ; 
what  was  the  breadth  of  the  street  ?  Ans.  60.734  feet. 

Ex.  5.  A  ladder  is  to  be  placed  so  as  to  reach  the  top  of  a 
wall  33.75  feet  high,  and  the  foot  of  it  can  not  be  set  nearer 
to  the  base  of  the  wall  than  18  feet ;  what  must  be  the  length 
of  the  ladder  !  Ans.  38.25  feet. 

Ex.  6.  The  base  of  an  isosceles  triangle  is  25  feet,  and  the 
sides  of  it  are  32.5  feet;  what  is  its  perpendicular  height? 

Ans.  30  feet. 

Ex.  7.  If  a  ladder  100  feet  in  length  was  set  upright  against 
a  vertical  wall,  and  then  set  out  at  its  foot  10  feet  from  the 
wall,  how  far  would  the  top  of  the  ladder  fall? 

Ans.  .50125  inches. 

Ex.  8.  The  width  of  the  wall  plates  of  a  house  is  48  feet, 
and  the  height  of  the  ridge  is  10  feet ;  what  must  be  the  length 
of  the  rafters  ?  Ans.  2Q>  feet. 

When  any  two  of  the  dimensions  of  a  triangle  and  one  of  the  correspond- 
ing dimensions  of  a  similar  figure  are  given,  and  it  is  required  to  find  the 
other  corresponding  dimensions  of  the  last  figure. 

Let  A  B  C,  a  b  c,  be  two  similar  triangles,  Figs.  8  and  9. 

C  -       Fig.  8. 

Fig.  9. 


B  Aba. 

nmAB:B  C  ::a  b  :  b  c,  or  a  b  :  b  c\\  A  B  :  B  C. 

Note. — The  same  proportion  holds  with  respect  to  the  similar  lineal 
parts  of  any  other  similar  figures,  whether  plane  or  solid. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  55 

Example. — The  shadow  of  a  cone  4  feet  in  length,  set  ver- 
tical, was  5  feet ;  at  the  same  time,  the  shadow  of  a  tree  was 
found  to  be  83  feet ;  what  was  the  height  of  the  tree,  both 
shadows  being  on  level  ground  ? 

ab:  k::AB:BC 
5  :  4:: 83  :  66| 
4 
5)332 
66$ /ee*. 
Ex.  2.  The  side  of  a  square  is  5  feet,  its  diagonal  7.071 
feet ;  what  will  be  the  side  of  a  square,  the  diagonal  of  which 
is  4  feet?  Ans.  2.828  feet. 

Ex.  3.  The  length  of  the  shadow  of  a  spire  is  151.5  feet, 
while  the  shadow  of  a  post  8  feet  high  is  6  feet ;  what  is  the 
height  of  the  spire  ?  Ans.  202  feet. 

To  ascertain  the  length  of  a  Side  when  the  Hypothenuse  of  a 
right-angled  Triangle  of  equal  sides  alone  is  given. 
Rule. — Divide  the  hypothenuse  by  1.414213,  and  the  quo- 
tient will  give  the  length  of  a  side. 

Or,  -   ; ,     *    ;  —the  length  of  a  side. 
'  1.414213  *      J 

Example. — The  hypothenuse  of  a  right-angled  triangle  is 
300  feet;  what  is  the  length  of  its  sides? 

300-^  1.414213  =  212.1320  feet. 

Ex.  2.  The  diagonal  of  a  square  is  28.28426  feet ;  what  is 
the  length  of  a  side  of  it  V  Ans.  20  feet. 

To  ascertain  the  Perpendicular  or  Height  of  a  Triangle  when  the 
+Base  and  Area  alone  are  given. 
Rule. — Divide  twice  the  area  of  the  triangle  by  its  base, 
and  the  result  is  the  length  of  the  perpendicular. 

Or,—  =  h. 

Example. — The  area  of  a  triangle  is  10  feet,  and  the  length 
of  its  base  5  ;  what  is  its  perpendicular  1 

10x2  =  20,  and  20-r-5  =  4/<W. 


56  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

Ex.  2.  The  base  of  an  isosceles  triangle  is  25  feet,  and*  its 
area  375  feet;  what  is  the  height  of  its  perpendicular? 

Ans.  30  feet. 

To  ascertain  the  Perpendicular  or  Height  of  a  Triangle  when  the 
two  Sides  and  the  Base  are  given  (Fig.  10). 

Rule. — As  the  base  is  to  the  sum  of  the  sides,  so  is  the  dif- 
ference of  the  sides  to  the  difference  of  the  divisions  of  the 
base.  Half  this  difference  being  added  to  or  subtracted  from 
half  the  base  will  give  the  two  divisions  thereof.  Hence,  as 
the  sides  and  their  opposite  division  of  the  base  constitute  a 
right-angled  triangle,  the  perpendicular  thereof  is  readily  found 
by  preceding  rules. 

L.     b  c+c  axb  cCOC  a     .    _       _ 

Or, 5 —b  dcod  a. 

b  a 

Or, — —a  d,  whence  -y/a  c2—ad2—dc. 

2  a  b  ' 

Fig.  10.  c 


Example. — The  three  sides  of  a  triangle,  a,  b,  c,  Fig.  10, 
are  9.928,  8,  and  5  feet ;  what  is  the  length  of  the  perpen- 
dicular on  the  longest  side  ? 

As  9.928  :  8 +5  : :  8  co5  :  3.928,  the  difference  of  the  divisions  of 
the  'base. 

Then  3.928-^2  =  1.964,  which,  added  to  -^p. =4.964  + 

1.964  =  6.928,  the  length  of  the  longest  division  of  the  base. 

Hence  we  have  a  right-angled  triangle  with  its  base  6.928,  and 
its  hypothenuse  8  ;  consequently,  its  remaining  side  or  perpendicular 
is  y'(82-6.9282)=4/^.      - 

Ex.  2.  The  three  sides  of  a  triangle  are  42,  40,  and  26  feet ; 
what  is  the  height  thereof?  Ans.  24  feet. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


57 


Centres  of  Gravity.  On  a  line  drawn  from  any  angle  to  the 
middle  of  the  opposite  side,  at  two  thirds  of  the  distance  from 
the  angle,  as  at  o,fig.  5. 


TRAPEZIUMS    AND    TRAPEZOIDS. 

Trapeziwn. 
Definition.      Quadrilaterals  having  unequal  sides. 

To  ascertain  the  Area  of  a  Trapezium  (Fig.  11). 

Rule. — Multiply  the  diagonal  a  c  by  the  sum  of  the  two 
perpendiculars  falling  upon  it  from  the  opposite  angles,  and 
half  the  product  is  the  area. 


~     acxb-t-d 

Or, =area. 


Fig.  11. 


Example. — The  diagonal  a  c,fig.  11,  is  125  feet,  and  the 
perpendiculars  b  and  d,  50  and  37  feet ;  what  is  the  area? 
125x50  +  37  =  10875,  and  10875  -f-  2 =5437.5  square  feet. 
Ex.  2.  What  is  the  area  of  a  plot  of  ground,  the  diagonal 
being  12.5  poles,  and  the  perpendiculars  50  and  37  poles? 

Ans.  543.75  poles. 
Ex.  3.  What  is  the  area  of  a  trapezium,  the  diagonal  being 
42  feet,  one  perpendicular  18  feet,  and  the  other  16? 

Ans.  714  feet. 

When  the  two  opposite  angles  are  supplements  to  each  other,  that 
is,  when  a  trapezium  can  be  inscribed  in  a  circle,  the  sum  of  its  op- 
posite angles  being  equal  to  two  right  angles,  or  180°. 

Rule. — From  half  the  sum  of  the  four  sides  subtract  each 
C  2 


58  MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

side  severally ;  then  multiply  the  four  remainders  continually 
together,  and  the  square  root  of  the  product  will  be  the  area. 

Example. — In  a  trapezium  the  sides  are  15,  13,  14,  and 
12,  and  the  diagonal  16  inches;  required  its  area,  its  opposite 
angles  being  supplements  to  each  other. 

15  +  13  +  14+12=54,  and  ^=27. 

27     27     27     27 
15     13     14     12 

12x14x13x15  =  32760,  and  -^32760  =  180.997,  Am. 

Centre  of  Gravity.  Draw  the  two  diagonals,  and  find  the 
centres  of  gravity  of  each  of  the  four  triangles  thus  formed ; 
join  each  opposite  pair  of  their  centres,  and  the  intersection 
of  the  two  lines  fs  the  centre  of  gravity. 

Trapezoid. 
Definition.     A  Quadrilateral  with  only  one  pair  of  opposite 


To  ascertain  the  Area  of  a  Trapezoid  (Fig.  12). 

Rule. — Multiply  the  sum  of  the  parallel  sides  a  b,  d  c,  by 
a  h,  the  perpendicular  distance  between  them,  and  half  the 
product  is  the  area. 
a  b-\-d  cxa  h 


Or, 


2 


Or, - —area,  s+/  representing  the  sides. 


b 


c  h  d 

Example. — The  parallel  sides  are  100  and  132  feet,  and 
the  distance  between  them  62.5  feet ;  what  is  the  area  ! 
100  +  132x62.5  =  14500,  and  14500-4-2=7250  square  feet. 


MENSURATION    OF   AREAS,  LINES,  AND    SURFACES.  59 

Ex.  2.  Required  the  area  in  feet,  the  distance  between  the 
parallel  sides  being  12.5  feet,  and  the  sides  being  20  and  26.4 
yards.  Ans.  870  feet. 

Ex.  3.  How  many  square  yards  are  there  in  a  trapezoid 
the  breadth  of  which  is  65  feet,  and  the  two  sides  28  and 
38.5  feet?  Ans.  240.1388. 

Ex.  4.  What  is  the  area  of  a  trapezoid  the  height  of  which 
is  54.25  feet,  and  the  sides  28  feet  1£  inches  and  30  feet  4^ 
inches'?        '  Ans.  1586.8125. 

Centre  of  Gravity.  On  a  line,  e,  joining  the  middle  points 
of  the  parallel  sides  a  b,  d  c,  the  distance  from 


e      (cd+2ab\ 

dc=-x( — -j- — r) 

3     \c  d-\-a  b  ) 


POLYGONS. 


Definition.  Plane  figures  having  more  than  four  sides,  and 
are  either  regular  or  irregular,  according  as  their  sides  or  angles 
are  equal  or  unequal,  and  they  are  named  from  the  number  of 
their  sides  and  angles.     Thus, 


A  Trig  on  (triangle)  has  3  sides. 

An  Octagon         lias    8  sides. 

Tetragon  (square)      4     " 

A  Nonagon                 9     " 

Pentagon                   5     " 

Decagon               10     " 

Hexagon                    6     " 

An  Undecagon           11     " 

Heptagon                   7     " 

A  Dodecagon            12    " 

fyc,  fyc,  fyc. 
All  of  which  being  composed  of  isosceles  triangles,  they  may  be  sim- 
ilarly measured. 

Regular  Polygons. 

To  ascertain  the  Area  of  a  regular  Polygon  (Fig.  13). 

Rule. — Multiply  the  length  of  a  side,  a  b,  by  the  perpen- 
dicular distance  to  the  centre  e  c,  and  the  product  multiplied 
by  the  number  of  sides  and  divided  by  2  will  be  the  area. 

Or, —area,  where  n  represents  the  number  of  sides. 


60  MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

Fig.  13.  a 


Example. — "What  is  the  area  of  a  pentagon,  the  side  a  b 
being  5  feet,  and  the  distance  c  e  4.25  feet? 
5  x  4.25  x5  (n)  =  106.25  =product  of  length  of  a  side,  the  dis- 
tance to  the  centre,  and  the  number  of  sides. 
Then  106.25^2  =  53.125  feet,  the  result  required. 
Ex.  2.  What  is  the  area  of  a  hexagon  when  its  side  is 
14.6  and  its  perpendicular  12.64  feet? 

A  ns.  553.632  square  feet. 
Ex.  3.  What  is  the  area  of  an  octagon  when  its  sides  are 
4.9705  and  its  perpendicular  6  feet?       Ans.  119.292  feet. 

To  ascertain  the  Area  of  a  Regular  Polygon  when  the  length  of  a 
Side  only  is  given. 
Eule. — Multiply  the  square  of  the  side  by  the  multiplier 
opposite  to  the  name  of  the  polygon  in  the  following  table, 
and  the  product  will  be  the  area. 


A. 

B. 

c. 

D. 

No.  of 
Sides. 

Name  of  Polygon. 

Area. 

Radius  of 

Circumscribed 

Circle. 

Length  of 
the  Side. 

Radius  of 

Circumscribing 

Circle. 

Radius  of 

Inscribed 

Circle. 

3 

Trigon 

0.4330 

2. 

1.7320 

.5773 

.2887 

4 

Tetragon 

1. 

1.414 

1.4142 

.7071 

.5 

5 

Pentagon 

1.7205 

1.238 

1.1756 

.8506 

.6882 

6 

Hexagon 

2.5981 

1.156 

1. 

1. 

.8660 

7 

Heptagon 

3.6339 

1.110 

.8677 

1.1524 

1.0383 

8 

Octagon 

4.8284 

1.083 

.7653 

1.3066 

1.2071 

9 

Nonagon 

6.1818 

1.064 

.6840 

1.4619 

1.3737 

10 

Decagon 

•  7.6942 

1.051 

.6180 

1.6180 

1.5388 

11 

Undecagon 

9.3656 

1.042 

.5634 

1.7747 

1.7028 

12 

Dodecagon 

11.1962 

1.037 

.5176 

1.9319 

1.8660 

Example. — What  is  the  area  of  a  square  when  the  length 
of  its  sides  is  7.0710678  inches? 

7.07106782=50,  and  50x1.  =50  inches,  Ans. 


MENSURATION   OF  AREAS,  LINES,  AND  SURFACES.  Gl 

Ex.  2.  "What  is  the  area  of  a  hexagon  when  the  length  of 
its  sides  is  5  inches'?  Ans.  64.9525  inches. 

Ex.  3.  Kequired  the  area  of  an  octagon,  the  length  of  its 
sides  being  3.8265  inches.  Ans.  70.698  inches. 

I 

To  ascertain  the  Radius  of  a  Circle  that  contains  a  given  Polygon 
when  the  length  of  a  Perpendicular  from  the  Centre  alone  is 
given. 

Rule. — Multiply  the  distance  from  the  centre  to  a  side  of 
the  polygon  by  the  unit  in  column  A,  and  the  product  will  be 
the  radius  of  a  circle  that  will  circumscribe  the  figure. 

Example. — What  is  the  radius  of  a  circle  that  contains  a 
hexagon,  the  distance  to  the  centre  being  4.33  inches'? 
4.33x1.156=5  inches. 

Ex.  2.  What  is  the  radius  of  a  circle  that  contains  an  oc- 
tagon, the  distance  from  a  side  to  the  centre  being  4.6168 
feet?  Ans.  5  feet. 

Ex.  3.  What  is  the  radius  of  a  circle  that  contains  a  square, 
the  distance  from  a  side  to  the  centre  being  3.5355  feet? 

Ans.  5  feet 

To  ascertain  the  length  of  a  Side  of  a  Polygon  that  is  contained  in 
a  given  Circle  when  the  radius  of  the  Circle  is  given. 

Rule. — Multiply  the  radius  of  the  circle  by  the  unit  in 
column  B,  and  the  product  will  be  the  length  of  the  side  of 
the  figure  which  the  circle  will  contain. 

Example.- — What  is  the  length  of  the  side  of  a  pentagon 
contained  in  a  circle  8.5  feet  in  diameter  ? 

-^-  =  4.25  radius,  and  4.25  x  1.1756  =  5  feet. 

Z 

Ex.  2.  What  is  the  length  of  the  side  of  a  hexagon  when 
the  diameter  of  the  circumscribing  circle  is  20  feet? 

Ans.  10  feet. 

Ex.  3.  What  is  the  length  of  the  side  of  an  octagon  when 
the  diameter  of  the  circle  is  10  feet?  Ans.  3.8265  feet. 


62  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

To  ascertain  the  Radius  of  a  Circumscribing  Circle  ivhen  the 
length  of  a  Side  is  given. 

Rule. — Multiply  the  length  of  a  side  of  the  polygon  by  the 
unit  in  column  C,  and  the  product  will  give  the  radius  of  the 
circumscribing  circle. 

Example. — What  is  the  radius  of  a  circle  that  will  contain 
a  hexagon,  a  side  being  5  inches  ? 

5x1=5  inches,  Ans. 
Ex.  2.  What  is  the  radius  of  a  circle  that  will  contain  a 
pentagon,  a  side  of  it  being  5.878  inches.         Ans.  5  inches. 

Ex.  3.  What  is  the  radius  of  a  circle  that  will  contain  an 
octagon,  when  each  of  its  sides  is  7.653  feet  ? 

Ans.  10  feet. 
Ex.  4.  Each  side  of  a  pentagon  is  required  to  be  9  feet ; 
what  are  the  radii  of  the  circumscribing  and  inscribed  circles  ? 
9  x. 8506  =  7. 6554  =  7  feet  7.8648   inches,  radius  of  circum- 
scribing circle. 
9  x. 6882=6.1938-6  feet  2.3256  inches,  radius  of  inscribed 

circle. 

To  ascertain  the  Radius  of  a  Circle  that  can  be  inscribed  in  a  given 
Polygon  when  the  length  of  a  Side  is  given. 

Rule. — Multiply  the  length  of  a  side  of  the  polygon  by  the 
unit  in  column  D,  and  the  product  is  the  radius  of  an  in- 
scribed circle. 

Example. — What  is  the  radius  of  the  circle  that  is  bound- 
ed by  a  hexagon,  its  sides  being  5  inches  ? 

5  x. 866  =  4.33  inches,  Ans. 

Ex.  2.  The  sides  of  an  octagon  are  8  inches ;  what  is  the 
radius  of  the  inscribed  circle?  Ans.  9.657  inches. 

Ex.  3.  The  sides  of  a  pentagon  are  5.878  inches ;  what  is 
the  radius  of  its  inscribed  circle?  Ans.  4.045  inches. 

To  ascertain  the  Length  of  a  Side  and  Radius  of  a  regular  Poly- 
gon when  the  Area  alone  is  given. 

Rule. — Multiply  the  square  root  of  the  area  of  the  polygon 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


63 


by  the  multiplier  in  column  E  of  the  following  table  for  the 
length  of  the  side  ;  by  the  multiplier  in  column  G  of  the  same 
table  for  the  radius  of  the  circumscribing  circle,  and  by  the 
multiplier  in  column  H,  also  in  the  same  table,  for  the  radius 
of  the  inscribed  circle  or  perpendicular. 


No.  of 
Sides. 

Name  of  Polygon. 

E. 

Length  of 
the  Side. 

G. 

Radius  of 

Circumscribing 

Circle. 

H. 

Radius  of 

Inscribed 

Circle 

Angle. 

Angle  of 
Polygon 

3 
4 
5 
6 

Trigon 
Tetragon 
Pentagon 
Hexagon 

1.5197 

1. 
.7624 
.6204 

.8774 

.7071 
.6485 
.6204 

.4387 
.5 

.5247 
.5373 

120° 
90 

72 
60 

60° 

90 
108 
120 

7 

8 

9 

10 

Heptagon 
Octagon 
Nonagon 
Decagon 

.5246 
.4551 
.4022 
.3605 

.6045 
.5946 
.5880 
.5833 

.5446 
.5493 
.5525 

.5548 

5125' 
45 
40 
36 

128f 
185 
140 
144 

11 
12 

Undecagon 
Dodecagon 

.3268 
.2989 

.5799 
.5774 

.5564 
.5577 

32  43' 
30 

147^ 
150 

Example. — The  area  of  a  square  (tetragon)  is  16  inches; 
what  is  the  length  of  its  side  ? 

-y/16z=4,  and  4  x  1=4  inches. 
Ex.  2.  The  area  of  an  octagon  is  70.698  yards;  what  is 
the  length  of  the  diameter  of  its  circumscribing  circle  ? 

Ans.  10  yards. 
Ex.  3.  The  area  of  a  square  is  50  inches ;  what  is  the 
length  of  the  radius  of  its  inscribed  circle? 

Ans.  3.5355  inches. 
Ex.  4.  The  area  of  a  hexagon  is  64.9525  inches;  what  is 
the  length  of  its  sides'?  Ans.  5  inches. 

Ex.  5.  The  area  of  a  decagon  is  144  inches ;  what  are  the 
lengths  of  its  sides,  and  of  the  radii  of  its*  circumscribing  and 
inscribed  circles'? 

•v/144  =  12,  and  12  x. 3605=4.326    inches,} 
y!44  =  12,  and  12  x.5833  =  6.9996      "      *>Ans. 
T/lU  =  12,and  12  x. 5548=6.6576      "     J 

Additional  uses  of  the  foregoing  Table. 

The  sixth  and  seventh  columns  of  the  table  will  greatly  fa 
cilitate  the  construction  of  these  figures  with  the  aid  of  a  sec- 


64 


MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 


tor.  Thus,  if  it  is  required  to  describe  an  octagon,  opposite 
to  it,  in  column  sixth,  is  45  ;  then,  with  the  chord  of  60  on  the 
sector  as  radius,  describe  a  circle,  taking  the  length  45  on  the 
same  line  of  the"  sector;  mark  this  distance  off  on  the  circum- 
ference, which,  being  repeated  around  the  circle,  will  give  the 
points  of  the  sides. 

The  seventh  column  gives  the  angle  which  any  two  adjoin- 
ing sides  of  the  respective  figures  make  with  each  other. 


REGULAR    BODIES. 

To  ascertain  the  Surface  or  Linear  Edge  of  any  Regular  Solid 

Body* 
Rule. — Multiply  the  square  of  the  linear  edge,  or  the  radius 
of  the  circumscribed  or  inscribed  circle,  by  the  units  in  the 
following  table,  under  the  head  of  the  dimension  used,  and 
the  product  will  be  the  surface  or  edge  required. 


Number 

Radius  of 

Radius  of 

of  sides. 

Names  of  figures. 

Surface. 

circum.  circle. 

inscribed  circle. 

4 

Tetrahedron 

1.73205 

1.63294 

4.89898 

6 

Hexahedron 

6. 

1.15470 

2. 

8 

Octahedron 

3.46410 

1.41421 

2.44949 

12 

Dodecahedron 

20.G4578 

.71364 

.89806 

20 

Icosahedron 

8.66025 

1.05146 

1.32317 

Example. — What  is  the  surface  of  a  hexahedron  or  cube 
having  sides  of  5  inches  ? 

52  x  6  =  25  x  6  =  150  inches,  Ans. 

¥%,  2.  What  is  the  linear  edge  of  a  hexahedron  circum- 
scribed by  a  circle  of  10  feet  radius?         Ans.  11.547  feet. 

Centre  of  Gravity.  In  all  regular  polygons  and  bodies  it  is 
at  their  geometrical  centre. 

Irregular  Polygons. 

Definition.     Figures  with  unequal  sides. 

To  ascertain  the  Area  of  an  Irregular  Polygon  {Fig.  14). 

Rule. — Draw  diagonals  to  divide  the  figures  into  triangles 

*  See  Appendix  (p.  258-61)  for  additional  rules  both  for  Polygons 
and  Regular  Bodies. 


MENSURATION    OF    AREAS,  LINES,  AND    SURFACES. 


65 


and  quadrilaterals :  find  the  areas  of  these  separately,  and  the 
sum  of  the  whole  is  the  area.* 

Note  or- ^b  ascertain  the  area  of  mixed  or  compound  figures,  or  such  as 
are  composed  of  rectilineal  and  curvilineal  figures  together,  compute  the  areas 
of  the  several  figures  of  which  the  whole  is  composed,  then  add  them  together, 
and  the  sum  will  be  the  area  of  the  compound  figure :  and  in  this  manner 
may  any  irregular  surface  or  field  of  land  be  measured,  by  dividing  it  into 
trapeziums  and  triangles,  and  computing  the  area  of  each  separately. 

a 
Fig.  U. 


Example. — What  is  the  content  of  Fig.  14? 
e  £=125  inches     be—  35.7  inches  ae=25  inches 

ag=  20      "         ec=130         "         and  d  i-20      " 
ebxa-g=125x20    =2500    -^-2  =  1250      ")         Ans. 
ebxb  c=125  x  35.7=4462.5-^-2  =  2231.25  I     4781.25 
ecxd  {=130x20    =2600    ^2  =  1300      Jj  ins. 

Ex.  2.  Required  the  area  of  the  irregular  figure  abode 
fga,  Fig.  15.  e 

#£=30.5        Fi2'15' 
gd=29 
fd=2A.S 
an-11.2 


*  Polygons  containing  one  or  more  re-entering  angles  are  called 
Re-entering,  as  Fig.  15.  The  term  re-entering  is  opposed  to  salient. 
It  is  a  property  of  a  salient  polygon  that  no  right  line  can  be  drawn  ex- 
ternal to  it  that  will  cut  its  perimeter  in  more  than  two  points,  while  in 
a  re-entering  polygon  such  line  may  cut  it  in  more  than  two  points. 


66  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

a  ni°  °  *9  b=U'2J~6 'x30.5  =  8.6x30.5  =  262.3=arga  of 

the  trapezium  a  b  c  g. 
'f  ltC  &  *9  ^=11t6'6 X  29=8.8  x29=255.2=area  of  the 

trapezium  g  c  df. 

f  dxe  P     24.8  x4 
*== — — z=99. 2=49. 6=area  of  the  triangle  f  d  e. 

Then  262.3  +  255.2  +  49.6  =  567.1  =  area  of  the  figure  required. 
Ex.  3.  Required  the  area  of  an  irregular  polygon. 
e  6=100  feet  e  c=110  feet 

ag—   18    "  b  c—   12    " 

a  e=  45    "  d  i=   15    " 

Ans.  2385  feet. 
Ex.  4.  In  a  pentangular  field,  beginning  at  the  south  side 
and  running  toward  the  east,  the  first  side  is  2735  links,  the 
second  3115,  the  third  2370,  the  fourth  2925,  and  the  fifth 
2220 ;  also  the  diagonal  from  the  first  to  the  third  is  3800 
links,  and  that  from  the  third  to  the  fifth  4010 ;  what  is  the 
area  of  the  field?  Ans.  117  ac.  2  ro.  39  po. 

Note. — As  this  figure  consists  of  three  triangles,  all  of  the  sides  of 
which  are  given,  by  calculating  their  areas  according  to  the  rule,  p.  51, 
the  sum  will  be  the  area  of  the  whole  figure  accurately,  without  draw- 
ing perpendiculars  from  the  angles  to  the  diagonals. 

The  same  thing  may  also  be  done  in  most  other  cases  of  this  kind. 


When  any  part  of  the  Figure  is  bounded  by  a  Curve,  the  Area 
may  be  found  as  follows : 

Rule. — Erect  any  number  of  perpendiculars  upon  the  base, 
at  equal  distances,  and  find  their  lengths. 

Add  the  lengths  of  the  perpendiculars  thus  found  together, 
and  their  sum  divided  by  their  number  will  give  the  mean 
breadth.  Multiply  the  mean  breadth  by  the  length  of  the 
base,  and  it  will  give  the  area  of  that  part  of  the  figure  re- 
quired. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


G7 


To  ascertain  the  Area  of  a  long  Irregular  Figure  {Fig.  16). 

Rule. — Take  the  breadth  at  several  places  and  at  equal 
distances  apart ;  add  them  together,  and  divide  their  sum  by 
the  number  of  breadths  for  the  mean  breadth ;  multiply  this 
by  the  length  of  the  figure,  and  the  product  is  the  area. 


Example. — What  is  the  area  o£fig.  16  in  square  feet? 

a  c=50  inches,  3  =   54  inches, 

1=52     "  hd-=   60     " 

2=48     "  cd=150     " 

50+52+48  +  54+60  =  264,  and  264^-5=52.8. 

Then  52.8  x  150  =  7920,  which-h- 144  =  55  square  feet,  Ans. 


To  ascertain  the  Centre  of  Gravity  of  any.  Plane  Figure. 

Rule. — Divide  it  into  triangles,  and  find  the  centre  of  gravity  of 
each ;  connect  two  centres  together,  and  find  their  common  centre ; 
then  connect  this  common  centre  and  the  centre  of  a  third,  and  find 
the  common  centre,  and  so  on,  always  connecting  the  last  found  com- 
mon centre  to  another  centre  till  the  whole  are  included,  and  the  last 
common  centre  will  be  that  which  is  required. 

Illustration.     Where  is  the  centre  of  gravity  of  Fig.  17  ? 

Fig.  17.  a     Fig.  19. 


Fig.  18. 


68  MENSURATION    OP  AREAS,  LINES,  AND    SURFACES. 

Fig.  20.     c  Fig.  21. 


The  point  •  represents  the  centre  of  gravity  of  each  triangle,  a  b  c, 
Fig.  18,  of  which  the  figure  is  composed. 

Connect  two  centres  of  gravity,  as  in  Fig.  19  ;  join  their  common  cen- 
tre to  the  centre  of  gravity  of  the  next  triangle,  as  in  Fig.  20 ;  join  their 
common  centre  o  to  the  centre  of  the  remaining  triangle,  Fig.  21,  and 
the  common  centre  of  this  last  connection  is  the  required  centre  of 
gravity  of  the  figure. 


CIRCLE. 

Definition.  A  plane  figure  bounded  by  a  true  curve,  called  its 
Circumference  or  Periphery. 

The  Diameter  of  a  Circle  is  a  right  line  drawn  through  its 
centre,  bounded  by  its  periphery. 

The  Radius  of  a  Circle  is  a  right  line  drawn  from  the  centre 
of  it  to  its  circumference. 

The  Circumference  of  a  Circle  is  assumed  to  be  divided  into 
360  equal  parts,  called  degrees;  each  degree  is  divided  into  60 
parts,  called  minutes ;  each  minute  into  60  parts,  called  seconds; 
and  each  second  into  60  parts,  called  thirds,  and  so  on. 

To  ascertain  the  Circumference  of  a  Circle  {Fig.  22). 

Rule. — Multiply  the  diameter  a  b  by  3.1416,*  and  the 
product  is  the  circumference. 

*  The  exact  proportion  of  the  diameter  of  a  circle  to  its  circumfer- 
ence has  never  yet  been  ascertained.  Nor  can  a  square  or  any  other 
right-lined  figure  be  found  that  shall  be  equal  to  a  given  circle.  This 
is  the  celebrated  problem  called  the  squaring  of  the  circle,  which  has 
exercised  the  abilities  of  mathematicians  for  ages,  and  been  the  occa- 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  69 

Or,  as  7  is  to  22,  so  is  the  diameter  to  the  circumference. 
Or,  as  113  is  to  355,  so  is  the  diameter  to  the  circumference. 


sion  of  so  many  disputes.  Several  persons  of  eminence,  have,  at  dif- 
ferent times,  pretended  that  they  had  discovered  the  exact  quadrature ; 
but  their  errors  have  soon  been  detected,  and  it  is  now  conceded  as  a 
thing  impracticable  of  attainment. 

Though  the  relation  between  the  diameter  and  circumference  can  not 
be  accurately  expressed,  it  may  yet  be  approximated  to  great  exactness. 
In  this  manner  was  the  problem  solved  by  Archimedes,  about  two  thou- 
sand years  ago,  who  discovered  the  proportion  to  be  nearly  as  7  to  22. 
This  he  effected  by  showing  that  the  perimeter  of  a  circumscribed  reg- 
ular polygon  of  192  sides  is  to  the  diameter  in  a  less  ratio  than  that 
of  3^-  to  1,  and  that  the  perimeter  of  an  inscribed  polygon  of  96  sides 
is  to  the  diameter  in  a  greater  ratio  than  that  of  3  -ff-  to  1,  and  from 
thence  inferred  the  ratio  above  mentioned,  as  may  be  seen  in  his  book 
De  Dimensione  Circuit. 

The  proportion  of  Vieta  and  Metius  is  that  of  113  to  355,  which  is 
something  more  than  the  former.  This  is  a  commodious  proportion ; 
for,- being  reduced  into  decimals,  it  is  correct  as  far  as  the  sixth  figure 
inclusively.  It  was  derived  from  the  pretended  quadrature  of  a  M.  Van 
Eick,  which  first  gave  rise  to  the  discovery. 

But  the  first  who  ascertained  this  ratio  to  any  great  degree  of  ex- 
actness was  Van  Ceulen,  a  Dutchman,  in  his  book  De  Circulo  et  Ad- 
scriptis.  He  found  that  if  the  diameter  of  a  circle  was  1,  the  circum- 
ference would  be  3.141592653589793238462643383279502884  nearly; 
which  is  exactly  true  to  36  places  of  decimals,  and  was  effected  by  the 
continual  bisection  of  an  arc  of  a  circle,  a  method  so  extremely  trouble- 
some and  laborious  that  it  must  have  cost  him  incredible  pains.  It  is 
said  to  have  been  thought  so  curious  a  performance,  that  the  numbers 
were  cut  on  his  tomb-stone  in  St.  Peter's  Church-yard  at  Leyden. 
This  last  number  has  since  been  confirmed  and  extended  to  double  the 
number  of  places  by  the  late  ingenious  Mr.  Abraham  Sharp,  of  Little 
Horton,  near  Bedford,  in  Yorkshire. 

But  since  the  invention  of  Eluxions,  and  the  Summation  of  Infinite 
Series,  there  have  been  several  methods  discovered  for  doing  the  same 
thing  with  much  more  ease  and  expedition.  The  late  Mr.  John  Machin, 
Professor  of  Astronomy  in  Gresham  College,  has  by  these  means  given 
a  quadrature  of  the  circle  which  is  true  to  100  places  of  decimals ;  and 
M.  de  Lagny,  M.  Euler,  &c,  have  carried  it  still  farther.  All  of  which 
proportions  are  so  Extremely  near  the  truth,  that,  except  the  ratio 
could  be  completely  obtained,  we  need  not  wish  for  a  greater  degree 
of  accuracy. — Bonntcastle. 


\ 

70  MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

Fig.  22. 


Example. — The  diameter  of  a  circle,  fig.  22,  is  1.25  inches ; 
what  is  its  circumference  ? 

1.25x3.1416  =  3.927  inches. 

Ex.  2.  The  diameter  of  a  circle  is  17  feet;  what  is  its  cir- 
cumference? Ans.  53. 4072  feet. 

To  ascertain  the  Diameter  of  a  Circle  (Fig.  22). 

Divide  the  circumference  by  3.1416,  and  the  quotient  is  the 
diameter. 

Or,  as  22  is  to  7,  so  is  the  circumference  to  the  diameter. 

To  ascertain  the  Area  of  a  Circle  (Fig.  22). 

Rule. — Multiply  the  square  of  the  diameter  by  .7854,  and 
the  product  is  the  area. 

Or,  multiply  the  squai*e  of  the  circumference  by  .07958. 

Or,  multiply  half  the  circumference  by  half  the  diameter. 

Or,  multiply  the  square  of  the  radius  by  3.1416. 

Or,  p  r2=area,  where  p  represents  the  ratio  of  the  circumfer- 
ence to  the  diameter,  and  r  the  radius. 

Example. — The  diameter  of  a  circle  is  8  inches ;  what  is 
the  area  of  it  ? 

82  or  8  x  8  =  64,  and  64  x  .7854=50.2656  inches. 

Ex.  2.  What  is  the  area  of  a  circle,  the  radius  being  39J 
yards?  .  Ans.  4839.8311  yards. 

Ex.  3.  What  is  the  area  of  a  circle  in  feet,  the  diameter 
being  15j  poles?  Ans.  3214.6587 /«?*. 

Ex.  4.  What  is  the  circumference  of  a  circle,  the  area  of 
which  is  an  acre?  Ans.  246  yds.,  1  ft.,  10^  in. 

Centre  of  Gravity.    Is  in  its  geometrical  centre.    Semicircle., 

4  r 

k — —distance  from  centre. 

'6.p  J 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


71 


USEFUL   FACTORS. 

In  which  p  represents  the  Circumference  of  a  Circle  the  Diameter 
of  which  is  1. 

Then 

p  =  3.1415926535897932384626  + 
2p  =  6.283185307179  + 
4 />  =  12.566370614359  + 
±p=  1.570796326794  + 
ip=  0.785398163397  + 


!>■? 

4.188790 

ip= 

.523598 

ip= 

.392699 

•fsP  = 

.261799 

sh>P= 

.008726 

l_ 

P 
2 

P 

.318309 
.636619 

4_ 

1.273239 

.079577 

1.772453 

.886226 

3.544907 


F97884 


.564189 


I 

4~"/> 

Vp  — 
Wp= 
2Vp= 

p 
p 

—  =114.591559 
P 
fp=t     2.094395 

5=     1.909859 
P 
36/>  =  113.097335 


In  which  the  Diameter  of  a  Circle  is  10. 

1.  Chord  of  the  arc  of  the  semicircle  =10. 

2.  Chord  of  half  the  arc  of  the  semicircle  ==  7.071067 

3.  Versed  sine  of  the  arc  of  the  semicircle  =  5. 

4.  Versed  sine  of  half  the  arc  of  the  semicircle  =  1.464466 

5.  Chord  of  half  the  arc  of  the  half  of  the  arc  of  the 

semicircle  -  3.82683 

6.  Half  the  chord  of  the  chord  of  half  the  arc  =  3.535533 

7.  Length  of  arc  of  semicircle  =  15.7075*63 

8.  Length  of  half  the  arc  ,of  the  semicircle  ==  7.853981 
Square  of  the  chord  of  half  the  arc  of  the  semicircle  (2.)     =50. 
Square  root  of  versed  sine  of  half  the  arc  (4.)  =  1.210151 
Square  of  versed  sine  of  half  the  arc  (4.)  =  2.144664 
Square  of  the  chord  of  half  the  arc  of  half  the  arc  of  the 

semicircle  (5.)  =14.64467 

Square  of  half  the  chord  of  the  chord  of  half  the  arc  (6.)    =12.5 

In  all  the  following  calculations,  p  is  taken  at  3.1416,  ip  at 
.7854,  \p  at  .5236,  and  whenever  the  decimal  figure  next  to 
the  one  last  taken  exceeds  5,  one  is  added.  Thus,  3.14159 
for  four  places  of  decimals  is  set  down  3.1416. 


72  MENSURATION    OF  AKEAS,  LINES,  AND    SURFACES. 

Circular  Arc. 
Definition.     A  part  of  the  Circumference  of  a  Circle. 
Fig.  23.        c 


The  Sine  of  an  Arc  is  a  line  running  from  one  extremity  of  an  arc 
perpendicular  to  a  diameter  joining  the  other  extremity,  as  a  d,Jig.  23. 

The  Sine  of  an  Angle  is  the  sine  of  the  arc  that  measures  that  angle. 

The  Versed  Sine  of  an  Arc  or  Angle  is  the  part  of  the  diameter  inter- 
cepted between  the  sine  and  the  arc,  as  d  b.  It  is  frequently  written 
versed  sine  of  half  the  arc. 

The  Complement  of  an  Arc  or  Angle  is  what,remains  after  subtracting 
the  angle  from  90°,  as  a  o  c,  Fig.  23. 

The  Supplement  of  an  Arc  or  Angle  is  what  remains  after  subtracting 
the  angle  from  180°,  as  a  o  e,  Fig.  23. 

TJie  Coversed  Sine  is  the  sine  of  the  complement  of  the  arc,  as  c  g, 
Fig.  23. 

Note. — For  other  illustrations  of  these  definitions,  see  figure, 
page  46. 


To  ascertain  the  Length  of  an  Arc  of  a  Circle  {Fig.  24)  when  the 
Number  of  Degrees  and  the  Radius  are  given. 

Rule  1.  As  180  is  to  the  number  of  degrees  in  the  arc, 
so  is  3.1416  the  radius  to  its  length. 

Rule  2.    Multiply   the    radius,   o  a,  of   the    circle    by 
.01745329,  and  the  product  by  the  degrees  in  the  arc. 

If  the  length  is  required  for  minutes,  multiply  the  radius 
by  .000290889;  if  for  seconds,  multiply  by  .000004848. 
(See  Table  of  length  of  Arcs,  page  130.) 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  73 

Fig.  24.     c 


Example. — The  number  of  degrees  in  an  arc,  oab,  are  90, 
and  the  radius,  o  b,  5  inches ;  what  is  the  length  of  the  arc? 
180  :  90  :: 5  x 3.1416  :  7.854  inches,  Ans. 
Ex.  2.  The  radius  of  an  arc  is  10,  and  the  measure  of  its 
angle  44°  30'  30" ;  what  is  the  length  of  the  arc  1 

10 x. 01745329  =  . 1745329,  which x 44=7.679447£,  the 

length  for  44°. 
10 x. 000290889  =  .00290889,  whichx 30  =  .0872667,  the 

length  for  30r. 
10 x. 000004848  =  .00004848,  wMx  30  =  .00 14544,  the 

length  for  30". 
Then,     7.67944761 

.0872667  1=7.7681687,  the  length  required. 
.0014544 J 
Or,  reduce  the  minutes  and  seconds  to  the  decimal  of  a  degree, 
and  multiply  by  it. 

See  Rule,  page  34.  30'  30//=.5083,  and  .1745329  from 
above  x  44.5083  =  7.768163. 

Ex.  3.  The  degrees  in  the  arc  of  a  circle  are  90,  and  the 
diameter  of  the  circle  is  20  feet ;  what  is  the  length  of  the 
arc?  Ans.  15.708. 

Ex.  4.  The  degrees  in  the  arc  of  a  circle  are  32°  38'  42", 
and  the  radius  of  it  is  25  inches ;  what  is  the  length  of  the  arc  % 
32o  38'  42"=32.645  degrees. 

Ans.  14.2441  inches. 
Ex.  5.  The  degrees  in  the  arc  of  a  circle  are  147°  21'  18", 
and  the  radius  of  its  circle  is  25  feet ;  what  is  the  length  of 
the  arc  ?  Ans.  64.2959  feet. 

D 


74  MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

When  the  Chord  of  half  the  Arc  and  the  Chord  of  the  Arc  are 
given. 

Rule. — From  8  times  the  chord,  a  b,  of  half  the  arc,  Fig. 

25,  subtract  the  chord,  a  c,  of  the  arc,  and  one  third  of  the 

remainder  will  be  the  length  nearly. 

k     8  c-c     , 

Or,  — - — ,  c  representing  the  chord  of  half  the  arc,  and  c 

the  chord  of  the  arc. 

Fig.  25.       b 


Example. — The  chord  of  half  the  arc,  a  b,  is  30  inches, 
and  the  chord  of  the  arc  48  ;  what  is  the  length  of  the  arc  ? 
30  x  8  =  240  =  8  times  the  chord  of  half  the  arc. 
240-48  =  192,  and  192-^-3=64  inches,  Ans. 
Ex.  2.  The  chord  of  half  the  arc  is  60  feet,  and  the  chord 
of  the  arc  96  ;  what  is  the  length  of  the  arc? 

Ans.  128  feet. 
Ex.  3.  The  chord  of  half  the  arc  is  17.67765  feet,  and  the 
chord  of  the  arc  25  ;  what  is  the  length  of  the  arc  ? 

Ans.  38.80706*/^. 
Ex.  4.  The  chord  of  half  the  arc  is  3.8268  inches,  and  the 
chord  of  the  arc  7.071 ;  what  is  the  length  of  the  arc1? 

Ans.  7.8478-f 

When  the  Chord  of  the  Arc  and  the  Versed  Sine  of  the  Arc  are 

given. 

Rule. — Multiply  the  square  root  of  the  sum  of  the  square 
of  the  chord,  a  c,  Fig.  25,  and  four  times  the  square  of  the 

*  The  exact  length  is  39.27  feet. 
+  The  exact  length  is  7.854  inches. 


MENSURATION    OF    AREAS,  LINES,  AND    SURFACES.         75 

versed  sine,  b  r  (equal  to  twice  the  chord  of  half  the  arc),  by- 
ten  times  the  square  of  the  versed  sine ;  divide  this  product 
by  the  sum  of  fifteen  times  the  square  of  the  chord  and  thirty- 
three  times  the  square  of  the  versed  sine ;  then  add  this  quo- 
tient to  twice  the  chord,  a  b,  of  half  the  arc,*  and  the  sum 
will  be  the  length  of  the  arc  very  nearly. 


v  c  +4uaXl0u2 
Or,  — — —  —  +  \/c2+4:  v2}  c  representing  the  chord,  and  v  the 

LO  C   ~v  OoV 

versed  sine. 

To  ascertain  the  Versed  Sine  when  the  Chord  of  Half  the  Arc 
and  Chord  of  the  Arc  are  given. 

Rule. — From  the  square  of  the  chord,  a  b,  of  half  the  arc,  sub- 
tract the  square  of  half  the  chord  of  the  arc  a  c  (—a  r2),  and  the 
square  root  of  the  remainder  is  the  versed  sine  (jb  r). 

Or  Vc'z  —  (c-r-'2,y  =  versed  sine. 

To  ascertain  the  Length  of  the  Chord  of  Half  the  Arc. 

Rule  1.  Divide  the  square  root  of  the  sum  of  the  square  of  the 
chord  of  the  arc  and  four  times  the  square  of  the  versed  sine  by  two, 
and  the  result  will  be  the  length.     . 
.     Or,  -\/c2-\-4:  v2-r-2=chord  of  half  the  arc. 

Rule  2.  From  the  sum  of  the  squares  of  half  the  chord  of  the  arc 
and  the  versed  sine,  take  the  square  root,  and  the  result  will  be  the 
length. 


7(i)v= 


Or,\J  [-)  -\-v2=zchord  of  half  the  arc. 

Rule  3.  Multiply  the  diameter  by  the  versed  sine,  and  the  square 
root  of  their  product  will  be  the  length. 
Or,  -\/dXv=chord  of  half  the  arc. 

To  ascertain  the  Versed  Sine  when  the  Chord  of  Half  the  Arc 
and  the  Diameter  are  given. 

Rule. — Divide  the  square  of  the  chord  of  half  the  arc  by  the 
diameter,  and  the  quotient  will  be  the  versed  sine. 
Or,  (c'2-7-d)=zversed  sine. 

*  The  square  root  of  the  sum  of  the  square  of  the  chord  and  four  times 
the  square  of  the  versed  sine  is  equal  to  twice  the  chord  of  half  the  arc. 


76  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

When  the  Chord  of  the  Arc  and  the  Diameter  are  given. 

Rule. — From  the  square  of  the  diameter  subtract  the  square  of 
the  chord  and  extract  the  square  root  of  the  remainder  ;  subtract  this 
root  from  the  diameter,  and  half  the  remainder  is  the  versed  sine. 

Or,  r —versed  sine. 

m 

When  the  Versed  Sine  is  greater  than  a  Semidiameter. 

Proceed  as  before,  but  add  the  square  root  of  the  remainder  (of  the 
squares  of  the  diameter  and  chord)  to  the  diameter,  and  half  the  sum 
is  the  versed  sine. 

Or, ~ =versed  sine. 

To  ascertain  the  Diameter. 

Rule  1 .  Divide  the  square  of  the  chord  of  half  the  arc  by  the 

versed  sine,  and  the  result  will  be  the  diameter. 

c'2 
Or,  — =diameter. 
v 

Rule  2.  Add  the  square  of  half  the  chord  of  the  arc  to  the  square 
of  the  versed  sine  ;  divide  this  sum  by  the  versed  sine,  and  the  re- 
sult will  give  the  diameter. 


(I)V 


Or, ^.diameter. 

v 

To  ascertain  the  Chord  of  the  Arc  when  the  Chord  of  Half  the 
Arc  and  the  Versed  Sine  are  given. 

Rule. — From  the  square  of  the  chord  of  half  the  arc  subtract  the 
square  of  the  versed  sine,  and  twice  the  square  root  of  the  remainder 
will  give  the  chord. 

Or,  VV2— v2)X2— chord  of  the  arc. 

Example. — The  chord  of  half  the  arc  is  60  inches,  and  the  versed 
sine  36  ;  what  is  the  length  of  the  chord  of  the  arc  1 

602— 362=2304,  and  -/2304  X  2=96,  Ans. 

When  the  Diameter  and  Versed  Sine  are  given. 

Rule. — Multiply  the  versed  sine  by  2,  and  subtract  the  product 
from  the  diameter ;  then  subtract  the  square  of  the  remainder  from 


MENSURATION    OF  AREAS,  LINES,  AND   SURFACES.  77 

the  square  of  the  diameter,  and  the  square  root  of  the  remainder  will 
give  the  chord. 

Or,  V(oX2— dY— d2—c. 

If  the  diameter  and  chord  of  half  the  arc  only  are  given,  find  the 
versed  sine,  as  per  rule,  p.  75,  then  proceed  as  above. 

Example. — The  diameter  of  a  circle  is  100  feet,  and  the  versed 
sine  of  half  the  arc  is  36  ;  what  is  the  length  of  the  chord  of  the  arc  1 
36  X  2— 100=28,  then  28s— 1002=9216,  and  -^9216=96,  Ans. 

Example. — The  chord  of  an  arc  is  80  inches,  and  its  versed 
sine  30 ;  what  is  the  length  of  the  arc? 

802  =  6400 —square  of  the  chord. 
302  =   900= square  of  the  versed  sine. 
-y/(6400-f-900x4)  — 100  =zsquare   root   of  the  square   of  the 
chord  and  four  times  the  square  of  the  versed  sine,  which  is, 
twice  the  chord  of  half  the  arc. 
Then,  100  x302  x  10  =  900000  —product  of  ten  times  the  square 
of  the  versed  sine  and  the  root  above  obtained. 
802x  15  =  96000«^15  times  the  square  of  the  chord. 
302x33  =  29700  =  33  times  the  square  of  the  versed  sine. 
125700 
.    .100x302xl0     900000  ,._       ___ 

l25700~  ^  125700  ?  '  ' 

or  twice  the  chord  of  half  the  a?  c =  107. 1599  feet. 
Ex.  2.  The  chord  of  an  arc  is  7.07107  inches,  and  the 
versed  sine  1.46447  ;  what  is  the  length  of  the  arc? 
7.07 1072       =50  —the  square  of  the  chord. 

1.464472  x  4=   8.5787  =4  times  the  square  of  the  versed  sine. 
V '58.57 87  =  7. 6536 = twice  the  chord  of  half  the 
arc. 
1.464472  X  10  =  21.4467  =  10  times  the  square  of  the  versed  sine. 

7.07l072x  15  =  750.  =15  times  the  square  of  the  chord. 

1.464472  X  33  =   70.7742  =33  times  the  square  of  the  versed  sine. 
820.7742 

Then>  7^S^P^+7-Q5d6=7'S53Q  inches- 


78  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  ■ 

Ex.  3.  The  chord  of  an  arc  is  96  feet,  and  the  versed  sine 

36  ;  what  is  the. length  of  the  arc?         Ans.  128.5918  feet. 

Ex.  4.  The  chord  of  an  arc  is  40  inches,  and  the  versed  sine 

15  ;  what  is  the  length  of  the  arc?  Ans.  53.58  inches. 

Ex.  5.  The  chord  of  an  arc  is  48  inches,  and  the  versed  sine 

18  ;  what  is  the  length  of  the  arc?        Ans.  64.2959  inches. 

Ex.  6.  The  chord  of  an  arc  is  60  inches,  and  .its  versed 

sine  10  ;  what  is  the  length  of  the  arc? 

Ans.  64.3493  inches. 

Ex.  7.  The  versed  sine  of  an  arc  is  2.5658,  and  the  chord 

31.6228  inches;  what  is  the  length  of  the  arc? 

Ans.  32.1747  inches. 

Ex.  8.  The  chord  of  an  arc  is  7.071,  the  chord  of  half  the 

arc  is  3.8268,  and  the  diameter  of  the  circle  10  inches;  what 

are  the  lengths  of  the  versed  sine  and  arc  ? 

By  Note,  page  75. 

3.82682— 14.6444 =square  of  half  the  chord  of  the  arc. 

7.071^-2=3.5356,  and  3.53562= 12.5 ^square  of  half  the 

chord  of  the  arc.     , 

Then,  -^14.6444  — 12.5  =  1  AG ±4:= versed  sine. 

Or,  by  preceding  rule,  page  75, 

3.82682     14.6444     ,  AnAA 

— — — == — — —  =  1Ao4:4:= versed  sine. 

Length  of  arc  by  rule,  p.  74,  Example  2,  p.  77,  7.8536  inches. 
Ex.  9.  The  chord  of  an  arc  is  96,  and  the  versed  sine  36 
inches ;  what  is  the  chord  of  half  the  arc  ?      Ans.  GO  inches. 

Ex.  10.  The  chord  of  an  arc  is  70.7107,  and  the  versed 
sine  14.6447  inches;  what  are  the  lengths  of 

The  arc?  Ans.    78.54    inches. 

The  chord  of  half  the  arc  ?  Ans.    38.268      " 

And  the  diameter  of  the  circle?     Ans.  100.  " 

When  the  Diameter  and  Versed  Sine  are  given. 

Eule — .Multiply  twice  the  chord  of  half  the  arc,  a  b,  Fig. 
25,  by  10  times  the  versed  sine,  b  r;  divide  the  product  by 
27  times  the  versed  sine  subtracted  from  60  times  the  diam- 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  79 

eter,  and  add  the  quotient  to  twice  the  chord  of  half  the  arc ; 
the  sum  will  be  the  length  of  the  arc  very  nearly. 

c  x  2  x  10  v 
Or,  — —z (-  2c',  d  representing  the  diameter. 

Example. — The  diameter  of  a  circle  is  100  feet,  and  the 
versed  sine  of  the  arc  25  ;  what  is  the  length  of  the  arc  1 

V25  x  100 =50= chord  of  half  the  arc. 
50  x  2  x25xl0=25000  =  fance  the  chord  of  half  the  arc  by  10 

times  the  versed  sine. 

100x60—25x27=5325  =  27  times  the  versed  sine  subtracted 

from  60  times  the  diameter. 


25000 


Then  =4.6948,  and  4.6948  +  50x2  =  104.6948  feet. 

Ex.  2-.  The  diameter  of  a  circle  is  10  inches,  and  the  versed 
sine  1.46447  ;  what  is  the  length  of  the  arc"? 

V  1.46447  X  10  =  3.8268=c/wrd  of  half  the  arc. 
3.8268  x  2  xl0xl.46447  =  112.0847  =  ^'ce  the  chord  of  half 

the  arc  by  10  times  the  versed  sine. 
10  x  60  —  1.46447  x  27=560.459  =  27  times  the  versed  sine  sub- 
tracted from  60  times  the  diameter. 
Then  112.0847 -f-560.459  =  . 19998,  and  .19998  added  to 
3.8268  x  2  {twice  the  chord  of  half  the  arc)  =  7.8536  inches. 

Ex.  3.  The  diameter  of  a  circle  is  10  inches,  and  the  versed 
sine  1.46447  ;  what  is  the  length  of  the  arc  ? 

Ans.  7.854  inches. 
Ex.  4.  The  diameter  of  a  circle  is  41.66  feet,  and  the  versed 
sine  15  ;  what  is  the  length  of  the  arc? 

Ans.  53.5799  feet. 
Ex.  5.  The  diameter  of  a  circle  is  200  feet,  and  the  versed 
sine  of  the  arc  72  ;  what  is  the  length  of  the  arc? 

Ans.  257.1837  feet. 
Ex.  6.  The  diameter  of  a  circle  is  80  feet,  and  the  chord 
of  half  the  arc  is  16.018  ;  what  is  the  versed  sine  and  what  is 
the  length  of  the  arc  ! 

Ans.  Versed  sine,  3.2072;  Length  of  arc,  32.2539  feet. 


80  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

Centre  of  Gravity.     Multiply  the  radius  of  the  circle  by  the 

chord  of  the  arc,  and  divide  the  product  by  the  length  of  the 

arc ;  the  quotient  is  the  distance  of  it  from  the  centre  of  the 

circle. 

r  x  c 
Or,  — —  =zdista?ice  from  the  centre  of  the  circle. 

EXAMPLES    UNDER   THE    SEVERAL   RULES. 

1.  The  degrees  in  the  arc  of  a  sector  are  30°  38/  42",  and 
the  radius  of  the  circle  50 ;  what  is  the  length  of  the  arc  ? 

Ans.  26.7429. 

2.  The  chord  of  an  arc  is  70.71067,  and  the  chord  of  half 
the  arc  38.268  ;  what  i&the  length  of  the  arc? 

Ans.  78.536. 

3.  The  chord  of  an  arc  is  48,  and  the  versed  sine  18  ;  what 
is  the  length  of  the  arc?  Ans.  64.2959. 

4.  The  chord  of  an  arc  is  50,  the  radius  40,  and  the  versed 
sine  8.775  ;  what  is  the  length  of  the  arc  ?        Ans.  54.0096. 

*5.  The  diameter  of  a  circle  is  10,  and  the  versed  sine  5  ; 
what  is  the  chord  of  the  arc,  the  chord  of  half  the  arc;  and 
the  length  of  the  arc  ? 

f  Chord  of  the  arc,  10. 

Ans.  -j  Chord  of  half  the  arc,    7.07107. 
\  Length  of  the  arc,         15.708. 

6.  The  diameter  of  a  circle  is  100,  and  the  chord  of  half 
the  arc  60 ;  what  is  the  versed  sine?  Ans.  36. 

7.  The  diameter  of  a  circle  is  100,  and  the  chord  of  the 
arc  60 ;  what  is  the  versed  sine,  the  chord  of  half  the  arc, 
and  the  length  of  the  arc  ? 

f  Versed  sine,  10. 

Ans.l  Chord  of  half  the  arc,  31.6228. 
I Length  of  the  arc,         64.3493. 

8.  The  chord  of  an  arc  is  96,  and  the  versed  sine  36  ;  what 
are  the  lengths  of  the  chord  of  half  the  aro,  the  diameter,  and 
the  arc?  (  Chord  of  half  the  arc,    60. 

Ans-i  Diameter  of  the  circle,  100. 

I  Length  of  the  arc,        128.5918. 


MENSURATION    OP  AREAS,  LINES,  AND    SURFACES.  81 

Proportions  of  the  Circle,  its  Equal,  Inscribed,  and  Circumscribed 
Squares. 

CIRCLE. 

1.  Diameter  X.8862)       0.-,      ^  , 

.  ,  )™~  >■  =  Side  of  an  equal  square. 

2.  Circumference X. 2821)  ^  H         * 

3.  Diameter  X  .7071" 


4.  Circumference  x  .2251  f  =  Side  of  the  inscribed  square. 

5.  Area  X  .9003, 

SQUARE. 

6.  A  Side  X  1.4142  ^Diameter  of  its  circumscribing  circle. 

7.  "  x  4.443  =  Circumference  of  its  circumscribing  circle. 

8.  "  X  1.128   ^Diameter  *     -  ,    .    , 
°'         it                                      '          ,            y  of  an  equal  circle. 

9.  "  X  3.545  =  Circumference)  * 

10.  Square  inches  X  1.273   = Round  inches. 

Note. — The  square  described  within  a  circle  is  one  half  the  area  of 
one  described  without  it. 

Sector  of  a  Circle. 

Definition.     A  part  of  a  Circle  bounded  by  an  arc  and  two 
radii. 

To  ascertain  the  Area  of  a  Sector  of  a  Circle  when  the  Degrees 
in  the  Arc  are  given  {Fig.  26). 

Kule. — As  360  is  to  the  number  of  degrees  in  the  sector, 
so  is  the  area  of  the  circle  of  which  the  sector  is  a  part  to  the 

area  of  the  sector. 

,  •» 

Or,  —area,  d  representing  the  degrees  in  the  arc  and  a 

ooO 

the  area  of  the  circle. 

Fig.  26.       b 

a  sZ^Z. .7^T>s.  c 


82  MENSURATION    OP  AREAS,  LINES,  AND    SURFACES. 

Example. — The  radius  of  a  circle,  o  a,  is  5  inches,  and  the 
number  oidegrees  of  the  sector  is  22°  30' ;  what  is  the  area? 
Area  of  a  circle  of  5  inches  radius  is  78.54  inches. 
Then,  as  360°  :  22°  30' : :  78.54  :  4.90875,  Ans. 
Ex.  2.  The  degrees  in  the  arc  of  a  sector  are  147°  21/  18", 
and  the  area  of  the  circle  is  1963.5  feet;  what  is  the  area  of 
the  sector? 

To  reduce  147°  21'  18"  to  a  decimal. 
21     18 
60  # 

60)1278 
60)213 
.355 
Then  .355  +  147  =  147.355. 

Ans.  803.6987/*?*. 
Ex.  3.  The  degrees  in  the  arc  of  a  sector  are  32°  38'  42", 
and  the  area  of  the  circle  is  1963.5  inches;  what  is  the  area 
of  the  sector?  Ans.  178.0512  inches. 

Ex.  4.  The  degrees  in  the  arc  of  a  sector  are  90°,  and  the 
area  of  the  circle  is  981.75  inches;  what  is  the  area  of  the 
sector?  •  Ans.  245.4375  inches. 

Ex.  5.  The  degrees  in. the  arc  of  a  sector  are  90°,  the  versed 
sine  1.46446  feet,  and  the  chord  of  half  the  arc  is  3.8268 ; 
what  is  its  area  ? 

3.82682  =  14.6446,  the  square  of  the  chord  of  half  the  arc. 

14.6446  -r- 1.46446  =  10  =  the  square  of  the  chord  of  half  the  arc 

-^-the  versed  sine,  which  is  the  diameter. 

102  x  .7854  =  78.54,  the  area  of  the  whole  circle. 

Then,  as  360°  :  90° ::  78.54  :  19.635  feet 

Note. — Divide  the  area  by  .7854,  and  the  square  root  of  the  quotient 
is  the  diameter  of  the  circle. 

Illustration.     The  area  of  a  circle  is  176.715  ;  what  is  its  diameter? 
176.715-^.7854=225,  and  ^225  =  15,  Ans. 


When  the  Length  of  the  Arc,  fyc,  are  given  {Fig.  26). 

Rule. — Multiply  the  length  of  the  arc,  a  c  b,  by  half  the 
length  of  the  radius,  a  o,  and  the  product  is  the  area. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  83 

T 

Or,  ay,-— area,  a  representing  the  arc,  and  r  the  radius. 

z 

Example. — The  length  of  the  arc  of  a  sector  is  7.854  inches, 

and  a  radius  of  it  is  5  ;  what  is  its  area  ? 
5 
7.854x^  =  7.854x2.5  =  19.635  inches. 

Ex.  2.  The  length  of  the  arc  of  a  sector  is  10.472  inches, 
and  a  radius  5  ;  what  is  its  area?  Ans.  26.18  inches. 

Ex.  3.  The  length  of  the  arc  of  a  sector  is  14.19  inches,  the 
diameter  of  the  circle  being  100  ;  what  is  its  area  ? 

Ans.  354.75  inches. 

Ex.  4.  The  radius  of  a  circle  is  25  feet,  and  the  versed  sine, 
b  r,  of  the  arc  of  a  sector  is  18 ;  what  is  the  area  of  the  sector? 

By  Eule,  page  78,  the  length  of  the  arc  is  64.2959. 

Ans.  803.69875  feet. 

Note. — If  the  diameter  or  a  radius  is  not  given,  see  Rules,  page  7G. 

Ex.  5.  What  is  the  area  of  a  sector  when  the  versed  sine 
of  its  arc  is  15,  and  the  chord  40  inches'? 


40-1-2  =400= square  of  half  the  chord  of  the  arc. 


152= 225=  square  of  the  versed  sine. 
lien,  - — ~- =41.666,  the  diameter,  and  — '- — =20.83 

15  L 

the  radius. 

20.833 


Length  of  arc  by  Mule,  page  78,  53.58,  and  53.58  x 

23 

558.116  inches. 

Ex.  6.  The  radius  of  a  circle  is  50  inches,  and  the  versed 
sine  of  the  arc  of  a  sector  of  it  is  25  ;  what  is  its  area  ? 

Ans.  2617.37  inches. 

Ex.  7.  The  diameter  of  a  circle  is  100  feet,  the  versed  sine 
of  the  arc  of  a  sector  is  36,  and  the  chord  of  half  the  arc  is 
60 ;  what  is  the  area  of  the  sector?        Ans.  3214.795  feet. 

Ex.  8.  The  length  of  the  arc  of  a  sector  is  104.6948  inches, 
the  chord  of  the  arc  is  86.6024,  and  the  versed  sine  of  it  is 
25  ;  what  is  the  area  of  the  sector  ? 

Ans.  2617.37  inches. 


84  MENSURATION    OF  AKEAS,  LINES,  AND    SURFACES. 

Centre  of  Gravity.  Multiply  twice  the  chord  of  the  arc  by 
the  radius  of  the  sector,  and  divide  their  product  by  three 
times  the  length  of  the  arc ;  the  quotient  is  the  distance  from 
the  centre  of  the  circle. 

2  c  r 
Or,  ———^distance  from  centre  of  circle  ;  r  representing  radi- 
us, and  I  the  length  of  the  arc. 

Example. — Where  is  the  centre  of  gravity  of  the  sector 
given  in  example  5  ? 

40  X  2  =  80  =  twice  the  chord  of  the  arc. 
80  x  20.833  =  1666. 64  =product  of  twice  the  chord  and  the 
radius. 

53.58x3  =  160.74=^ree  times  the  length  of  the  arc. 
Then,  1666.64—160.74= 10.369,  the  distance  from  the  centre 
of  the  circle. 

Segment  of  a  Circle. 
Definition.     A  part  of  a  circle  hounded  by  an  arc  and  a  chord. 

To  ascertain  the  Area  of  a  Segment  of  a  Circle,  Fig.  27,  when 
the  Chord  and  Versed  Sine  of  the  Arc,  and  Radius  or  Diam- 
eter of  the  Circle  are  given. 

When  the  Segment  is  less  than  a  Semicircle,  as  ab  c,  Fig.  27. 

Rule. — Find  the  area  of  the  sector  having  the  same  arc  as 
the  segment ;  then  find  the  area  of  the  triangle  formed  by  the 
chord  of  the  segment  and  the  radii  of  the  sector,  and  the  dif- 
ference of  these  areas  will  be  the  area  required. 

Note. — Subtract  the  versed  sine  from  the  radius;  multiply  the  re- 
mainder by  one  half  of  the  chord  of  the  arc,  and  the  product  will  be 
the  area  of  the  triangle. 

Or,  a— a' —area  of  segment;  a  representing  area  of  the  sector, 
and  a'  the  area  of  the  triangle. 

When  the  Segment  is  greater  than  a  Semicircle,  as  a  e  c,  Fig.  27. 
Rule. — Find,  by  the  preceding  rule,  the  area  of  the  lesser 
portion  of  the  circle,  a  b  c ;  subtract  it  from  the  area  of  the 
whole  circle,  and  the  remainder  is  the  area  required. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  85 

Or,  c—c'=area  of  segment;  c  representing  area  of  circle,  and 
c'  area  of  the  lesser  portion. 

(See  Table  of  Areas,  page  134.) 
Fig.  27.         b 


Example. — The  chord,  a  c,  Fig.  27,  is  14.142,  the  diameter, 

b  e,  is  20,  and  the  versed  sine,  b  d,  is  2.929  inches ;  what  is  the 

area  of  the  segment  ? 

14.142 
By  Rule,  page  75,  — - —  =  7.071  =  half  the  chord  of  the  arc. 

m 

V7.0712+2.9292=7.654  =  ^e  square  root  of  the  sum  of  the 
squares  of  half  the  chord  of  the  arc  and  versed  sine,  which  is 
the  chord  a  b  of  half  the  arc  a  b  c. 

By  Rule,  page  78. 

7.654  x  2  x  10  x  2.929  =  448.371  =  twice  the  chord  of  half  the 
arc  by  10  times  the  versed  sine. 

20  x  60  —  2.929  X  27  =  1120.917  =  00  times  the  diameter  sub- 
tracted from  27  times  the  versed  sine. 


Then,  448.37 1-r- 1120,917 =.400,  and  .400  added  to  7.654x2 
{twice  the  chord  of  half  the  arc)  =  15.708  inches,  the  length  of 

the  arc.     By  Rule,  p.  82, 15.708  x  —= 78.54  =  the  arc  multi- 

plied  by  half  the  length  of  radius,  which  is  the  area  of  the  sector. 

10  —  2.929  =  7.071  =  ^6  versed  sine  subtracted  from  a  radius, 

14.142 
which  is  the  height  of  the  triangle  a  o  c,  and  7.071  x  — ~ — 

=  50  =  araz  of  the  triangle. 
Consequently,      78.54—50  =  28.54  inches. 


86  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

Ex.  2.  The  chord  of  the  arc  of  a  segment  is  86.6024,  the 
versed  sine  25,  and  the  radius  50  feet;  what  is  the  area  of 
the  segment  ?  Ans.  1534.84:  feet. 

Ex.  3.  The  chord  of  the  arc  of  a  segment  is  28  feet,  the 
diameter  of  the  circle  100,  and  the  versed  sine  of  the  arc  2  ; 
what  is  the  area  of  the  segment?  Ans.  37.4852  feet. 

Ex.  4.  The  diameter  of  a  circle  is  50  feet,  the  chord  of  the 
arc  of  a  segment  of  it  is  30,  and  its  versed  sine  5 ;  what  is 
the  length  of  the  arc,  and  what  the  area  of  the  segment  ? 

Ans.  Arc,  32.17 46  feet;  Segment,  102.183  feet. 

Ex.  5.  The  chord  of  a  segment  is  56  inches,  its  versed  sine 
4,  and  the  radius  of  the  circle  200  inches ;  what  is  its  area1?* 

Ans.  149.941  inches. 

When  the  Chords  of  the  Arc,  and  of  the  half  of  the  Arc,  and 
the  Versed  Sine  are  given. 

Rule. — To  the  chord,  a  c,  of  the  whole  arc,  add  the  chord, 
a  b,  of  half  the  arc  and  one  third  of  it  more ;  multiply  this 
sum  by  the  versed  sine,  b  d,  and  this  product,  multiplied  by 
.40426,  will  give  the  area  nearly. 
c' 
Or,  c-\-c  -\-—  xv  x  .40426= a?*ea  nearly, 
o 

Example. — The  chord  of  a  segment  is  28  feet,  the  chord 
of  half  the  arc  is  15,  and  the  versed  sine  6 ;  what  is  the  area 
of  the  segment  ? 

28  +  15-j-  — =4:8  =  the  chord  of  the  arc  added  to  the  chord  of 
o 

half  the  arc  and  ^  of  it  more. 

48  x  6  =  288— product  of  above  sum  and  the  versed  sine. 

Then,  288  x  .40426  =  116.427  feet,  the  area  required. 

Ex.  2.  The  chord  of  a  segment  is  40  feet,  its  versed  sine 

10,  and  the  chord  of  half  the  arc  is  22.36  ;  what  is  its  area? 

Ans.  282.226  feet. 

Ex.  3.  What  is  the  area  of  a  segment,  its  half  chord  being 

14,  the  chord  of  half  its  arc  14.142,  and  its  versed  sine  2  yards? 

Ans.  37.884  yards. 


MENSURATION   OF   AREAS,  LINES,  AND    SURFACES.  87 

Ex.  4.  The  chord  of  a  segment  is  150  feet,  the  chord  of 
half  the  arc  is  106.066,  and  the  versed  sine  75 ;  what  is  the 
area?  Ans.  8835.739  feet* 

When  the  Chord  of  the  Arc  (or  Segment)  and  the  Versed  Sine 
only  are  given. 

Rule. — Find  the  chord  of  half  the  arc,  and  proceed  as  be- 
fore. 

Example. — The  chord  of  the  arc  of  a  segment  is  28  yards, 

and  its  versed  sine  2  ;  what  is  the  length  of  the  chord  of  half 

the  arc,  and  what  the  area  of  the  segment  in  feet  ? 

28 

—=14,  and  142-f-22=200=swm  of  square  of  half  the  chord 

and  the  versed  sine. 
y200  =  14. 142 13 =square  root  of  preceding  sum = chord  of 
half  the  arc. 

Area  of  segment  113.6525  feet. 
Ex.  2.  The  chord  of  a  segment  is  48  inches,  its  versed  sine 
32,  and  the  diameter  of  the  circle  50 ;  what  is  its  area? 

The  versed  sine  being  greater  than  half  the  diameter,  the  segment 
is  consequently  greater  than  a  semicircle. 

Area  of  circle,  502  x  .7854=1963.5 
Area  of  lesser  portion,  versed  sine  18.=   640.3478 

Ans.  1323.1522  inches. 
Ex.  3.  The  chord  of  an  arc  is  86.6024,  and  its  versed  sine 
25  feet ;  what  is  the  area  of  the  segment  in  feet  and  inches? 

Ans.  1549  feet,  .156  inches. 
Centre  of  Gravity.     Divide  the  cube  of  the  chord  of  the  seg- 
ment by  twelve  times  the  area,  and  the  quotient  is  the  distance 
from  the  centre  of  the  circle. 

c3 
Or,  — —  —  d,  when  c  represents  the  chord  of  the  segment,  a  the 

area,  and  d  the  distance  from  the  centre  of  the  circle. 

Example. — The  chord  of  a  segment  is  14.14213,  its  radius 
10,  and  its  area  28.53 ;  where  is  its  centre  of  gravity? 
*  The  exact  area  is  8835.729. 


88  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

14.142133  =  2828.426  =  cwfo  of  the  chord. 
28.53  x  12  =  342.36  =  12  times  the  area. 
Then,  2828.426  -4-  342. 36 =8.261  from  the  centre  of  the  circle, 
and  10  —  8. 26 1  =  1.7  39  from  the  base  of  the  segment. 

Sphere. 

Definition.     A  figure,  the  surface  of  which  is  at  a  uniform 
distance  from  the  centre. 

To  ascertain  the  Convex  Surface  of  a  Sphere  (Fig.  28). 
Fig.  28.  c 


Rule. — Multiply  the  diameter,  a  b,  by  the  circumference, 
abed,  and  the  product  will  give  the  surface  required. 

Or,  dxc= surf  ace,  d  representing  diameter,  and  c  the  circum- 
ference. 

Or,  4  p  r2=surface*     Or,  p  d2  —  surface. 

Example. — What  is  the  surface  of  a  sphere  of  10  inches 
diameter u? 

10x31.416  =  314.16  inches. 

Ex.  2.  The  diameter  of  a  sphere  is  17  inches;  what  is  the 
surface  of  it  in  feet?  Ans.  6.305  square  feet. 

Ex.  3.  If  the  circumference  of  a  sphere  is  50.2656  inches, 
what  is  its  surface  in  feet?  Ans.  5. 585  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

*  p  or  7T  represents  in  this,  and  in  all  cases  where  it  is  used,  the  ra- 
tio of  the  circumference  of  a  circle  to  its  diameter,  or  3.1416. 


f 


MENSURATION    OF   AREAS,  LINES,  AND    SURFACES.  89 

Segment  of  a  Sphere. 
Definition.     A  section  of  a  sphere. 

To  ascertain  the  Surface  of  a  Segment  of  a  Sphere,  Fig.  29. 

Rule. — Multiply  the  height,  b  o,  by  the  circumference  of 
the  sphere,  and  the  product,  added  to  the  area  of  the  base, 
a  o  c,  is  the  surface  required. 

Or,  hxc-\-b=z  surf  ace,  when  h  represents  the  height,  c  the  cir- 
cumference of  the  sphere,  and  b  area  of  base. 

Or,  2prh=zconvex  surface  alone. 

Fig.  29.  b 


Example. — The  height,  b  o,  of  a  segment,  a  be,  is  36  inches, 
and  the  diameter,  b  e,  of  the  sphere,  100 ;  what  is  the  convex 
surface,  and  what  the  whole  surface  % 


36  x  100  x  3.1416  =  11309.76 —height  of  segment  multiplied  by 
the  circumference  of  the  sphere. 

Then,  to  ascertain  the  area  of  the  base.     The  diameter  and  versed 
sine  being  given  ;  the  diameter  of  the  base  of  the  segment,  being 
equal  to  the  chord  of  the  arc,  is,  by  rule,  page  76, 
36x2-100=28. 

V282-1002  =  96. 
962x. 7854  =  7238.2464 ^convex  surface,  and  7238.2464  + 
11309.76  =  18548.0064 —convex  surface  added  to  area  of 
base^the  whole  surface. 

Note. — When  the  convex  surface  of  a  figure  alone  is  required,  the 
area  or  areas  of  the  base  or  ends  must  be  omitted. 


90  MENSURATION   OF   AREAS,  LINES,  AND    SURFACES. 

Ex.  2.  The  height  of  a  segment  is  10,  and  the  diameter  of  the 
sphere  100  feet ;  what  is  the  surface"?       Ans.  5969.04  feet. 

Ex.  3.  The  diameter  of  a  sphere  is  200  inches,  and  the 
height  of  a  segment  of  it  is  1  foot  8  inches ;  what  is  its  surface 
in  feet !  Ans.  162.4733  feet. 

When  the  diameter  of  the  Base  of  the  Segment  and  the  Height 
of  it  are  alone  given. 
Rule. — Add  the  square  of  half  the  diameter  of  the  base  to 
the  square  of  the  height ;  divide  this  sum  by  the  height,  and 
the  result  will  give  the  diameter  of  the  sphere. 

2 

Or,  d  -^-2  -{-h2-i-h— diameter. 

Example. — The  semi-diameter  of  the  base  of  a  segment  of 
a  sphere  is  48  feet,  and  the  height  of  it  is  36 ;  what  is  the 
surface  of  the  segment  in  square  yards  I 

Ans.  2060.8896  yards. 

Centre  of  Gravity  of  Convex  Surface.  At  the  middle  of  its 
height. 

Spherical  Zone  {or  Frustrum  of  a  Sphere). 

Definition.  The  part  of  a  sphere  included  between  two  paral- 
lel chords. 

To  ascertain  the  Surface  of  a  Spherical  Zone,  Fig.  30. 
Fig.  30,      * 

\ 
\ 


Rule. — Multiply  the  height,  c  g,  by  the  circumference  of 
the  sphere,  and  the  product  added  to  the  area  of  the  two  ends 
is  the  surface  required. 

Or,  h  x  c  -f  a -f a' = surface. 

Or,  2  p  x  r  x'h— convex  surface. 


MENSURATION   OF  AREAS,  LINES,  AND    SURFACES.  91 

Example. — The  diameter  of  a  sphere,  a  b,  from  which  a 
segment  is  cut,  is  25  inches,  and  the  height  of  it,  c  g,  is  8 ; 
what  is  its  convex  surface? 
25x3.1416x8  =  628.32  =  height x circumference   of  sphere  = 

convex  surface. 

Ex.  2.  The  height  of  a  zone  is  36  inches,  and  the  radius 
of  the  sphere  is  50  inches ;  what  is  its  convex  surface  ? 

Ans.  11309.76  inches. 

When  the  Diameter  of  the  Sphere  is  not  given. 

Multiply  the  mean  length  of  the  two  chords  by  half  their  difference, 
divide  this  product  by  the  breadth  of  the  zone,  and  to  the  quotient  add 
the  breadth.  To  the  square  of  this  sum  add  the  square  of  the  lesser 
chord,  and  the  square  root  of  their  sum  will  be  the  diameter  of  the 
circle. 

Ex.  3.  The  greater  and  lesser  chords  of  a  segment  of  a 
sphere  are  96  and  60,  and  the  height  of  the  segment  is  26 ; 
what  is  its  convex  surface  and  what  its  surface  j 

Convex  surface,    8168.160")    \ 
Surface,    '  18233.846  j 

Centre  of  Gravity.     At  the  middle  of  its  height. 

Spheroids  or  Ellipsoids. 

Definition.  Figures  generated  by  the  revolution  of  a  semi- 
ellipse  about  one  of  its  diameters. 

When  the  revolution  is  about  the  transverse  diameter  they  are 
Prolate,  and  when  it  is  about  the  conjugate  they  are  Oblate. 

To  ascertain  the  Surface' of  a  Spheroid  {Fig.  31). 
When  the  Spheroid  is  Prolate. 
Rule. — Square  the  diameters,  a  b  and  c  d,  and  multiply  the 
square  root  of  half  their  sum  by  3.1416,  and  this  product  by 
the  conjugate  diameter. 

a"2  4-  d"2. 
Or,  <y/ — - x3.1416xd=:  surface,  d  representing  conjugate 

diameter. 


92  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

Fig.  31.       c 


Example. — A  prolate  spheroid  has  diameters  of  10  and  14 

inches ;  what  is  its  surface  ? 

102-{-142=296  = sum  of  squares  of  diameters. 

296  +  2  =  148,  and  ^/ 148  =  12.165  5= square  root  of  half  the 
sum  of  the  squares  of  the  diameters.  -^ 

12.1655  x  3.1416  x  10=382.191  =/?rato  of  root   above  ob- 
tained X  3.1416,  and  that  product  by  the  conjugate  diameter. 
Ex.  2.  A  prolate  spheroid  has  diameters  of  16   and  22 

inches;  what  is  its  surface?  Ans.  966.879  inches. 

When  the  Spheroid  is  Oblate. 

Rule. — Square  the  diameters,  a  b  and  c  d,  and  multiply 
the  square  root  of  half  their  sum  by  3.1416,  and  this  product 
by  the  transverse  diameter. 

d2-\-d/2  ,  , 

Or, -y/ - — X  3.1416  x  d' =z  surf  ace,  d'  representing  trans- 

verse  diameter. 

Example. — An  oblate  spheroid  has  diameters  of  14  and 

10  inches;  what  is  its  surface? 

142-f-102=296  = sum  of  squares  of  diameters. 

296  +  2  =  148,  and  ^148  =  12. 1655= square  root  of  half  the 
sum  of  the  squares  of  the  diameters. 

12.1655  x  3.1416  x  14  =  535.0679=proto  of  root  above  ob- 
tained x  3.1416,  and  that  product  by  the  transverse  diameter. 
Ex.  2.  An   oblate  spheroid  has  diameters  of  22   and  16 

inches;  what  is  its  surface?  Ans.  1329.4585  inches. 

Centre  of  Gravity.     Is  in  their  geometrical  centres. 
Note. — For  centre  of  gravity  of  semi-spheroids,  see  Appendix,  p.  281. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.  93 

To  ascertain  the  Convex  Surface  of  a  Segment  ofa  Spheroid  (Fig.  32), 

Rule. — Square  the  diameters,  and  take  the  square  root  of 
half  their  sum.  Then,  as  the  diameter  from  which  the  seg- 
ment is  cut,  is  to  this  root,  so  is  the  height  of  the  segment  to 
the  proportionate  height  required. 

Multiply  the  product  of  the  other  diameter  and  3.1416  by 
the  proportionate  height  of  the  segment,  and  this  last  product 
will  give  the  surface  required. 

Or,  5 ^ x  cT  or  d  x  3. 14 1 6  =  surface. 

a  or  a 

Fig.  32.  c 


Example. — The  height,  a  o,  of  a  segment,  e  f,  of  a  prolate 
spheroid,  Fig.  32,  is  4  inches,  the  diameters  being  10  and  14 
inches  ;  what  is  the  convex  surface  of  it  ? 
Square  root  of  half  the  sum  of  the  squares  of  the  diameters,  as 

by  previous  examples,  page  92,  12.1655. 
Then,  14  :  12.1655  ::4  :  3A7  58  =  height  of  segment,  proportion- 
ate to  the  mean  of  the  diameters. 
10  X  3.1416  X  3.4758  =  109.1957  =  remaining  diameter  x 

3.1416,  and  again  by  proportionate  height  of  segment. 
Ex.  2.  The  height  of  a  segment  of  a  prolate  spheroid,  Fig. 
32,  is  6  inches,  the  diameters  being  15  and  21  inches;  what 
is  the  convex  surface  of  it?  Ans.  252.9663  inches. 


94 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


Ex.  3.  The  height^  c  o,  of  a  segment,  e  f  of  an  oblate 
spheroid,  Fig.  33,  is  5  inches,  the  diameters  being  14  and  10 ; 
what  is  the  convex  surface?  Ans.  267.5339  inches. 

To  ascertain  the  Convex  Surface  of  a  Frustrum  or  Zone  of  a 
Spheroid  {Fig.  34). 

Rule. — Proceed  as  by  previous  rule,  page  93,  for  the  surface 
of  a  segment,  and  obtain  the  proportionate  height  of  the  frus- 
trum. Then  multiply  the  product  of  the  diameter  parallel  to 
the  base  of  the  frustrum  and  3.1416  by  the  proportionate  height 
of  the  frustrum,  and  it  will  give  the  surface  required. 

Or,  d  or  oV  x  3.1416  x  h=surface. 
Fig.  Si.       c 


Example. — The  middle  frustrum,  o  <?,  of  a  prolate  spheroid, 
Fig.  34,  is  6  inches,  the  diameters  of  the  spheroid  being  10 
and  14  inches ;  what  is  its  convex  surface  ? 

Mean  diameter,  as  per  example,  page  92,  is  12.1655. 
'Diameter  parallel  to  base  of  frustrum  is  10. 
As  14  :  12.J.655  ::  6  :  5. ,2138= proportionate  height  of  frustrum. 
10x3.1416  x5.2138  =  163.7967=swr/ace. 
Fig.  35.       c 


Ex.  2.  The  middle  frustrum  of  an  oblate  spheroid,  o  e9  Fig. 
35,  is  2  inches  in  height,  the  diameters  of  the  spheroid,  as  in 


MENSURATION    OF    AREAS,  LINES,  AND    SURFACES.  95 

the  preceding  examples,  being  10  and  14 ;  what  is  its  convex 
surface?  Aris.  107-0136  inches. 

Centre  of  Gravity.     Zone.     Is  in  its  geometrical  centre. 

Segment  and  Frustrum  of  Spheroid.    See  Appendix,  p.  28 1. 

Circular  Zone. 
Definition.     A  part  of  a  circle  included  between  two  parallel 
chords. 

To  ascertain  the  Area  of  a  Circular^  Zone  (Fig.  36). 

Rule. — To  the  area  of  the  trapezoid,  a  b  c  d,  or  of  the  par- 
illelogram,  a  h  c  g,  as  the  case  may  be,  add  the  area  of  the 
segments,  a  b,  c  d,  or  a  h,  c  g,  and  the  sum  is  the  area. 

Or,  subtract  the  areas  of  the  segments  a  i  c,  h  k  g,  from  the 
area  of  the  circle. 

Or,  a-{-a'z=zS,  a  representing  area  of  trapezoid,  or  parallelo- 
gram, and  a'  area  of  segments. 

(See  Table  of  Areas  of  Zones,  page  130.) 
Fig.  36.  i 


When  the  Diameter  of  the  Circle  i$  not  given. 

Multiply  the  mean  length  of  the  two  chords  by  half  their  difference ; 
divide  this  product  by  the  breadth  of  the  zone,  and  to  the  quotient  add 
the  breadth. 

To  the  square  of  this  sum  add  the  square  of  the  lesser  chord,  and 
the  square  root  of  their  sum  will  be  the  diameter  of  the  circle. 

Example. — The  greater  chord,  b  d,  is  96  inches,  the  lesser, 
a  c,  is  60,  and  the  breadth  of  the  zone,  a  e,  is  26.;  what  is  its 
area? 


96  MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

=78= mean  length  of  chords. =18= half  their 

difference. 
78x18 


54:=product  of  chords  and  their  difference,  divided  by 
*o 

the  breadth  of  the  zone. 

54  +  26  =  80=swm  of  above  quotient  and  breadth  of  zone. 

802  +  602  =  10000  =;  sum  of  square  of  above  sum  and  lesser  chord. 

Then,  -y/1 0000  =  100  =  diameter  required. 

— ±— ^=78,  and  78  X  2Q>-202S-area  of  trapezoid. 


To  ascertain  the  Area  of  the  Segments. 

It  is  necessary,  first,  to  ascertain  the  chord  of  their  arcs  ;  second,  the 
versed  sine  of  their  arcs. 

To  ascertain  the  Chord.  The  breadth  of  the  zone  is  the  perpendic- 
ular, a  e,  of  the  triangle,  of  which  either  chord,  a  b,  c  d,  is  the  hypothe- 
nuse.  Further,  half  the  difference  of  the  chords  a  c  and  b  d  of  the  zone 
is  the  length  of  the  base,  b  e,  of  this  triangle. 

Hence,  having  the  base  and  the  perpendicular,  the  hypothenuse  or 
chord  of  the  arc  of  the  segment  is  readily  found. 

Thus,  2Q=breadtk  of  the  zone  or  perpendicular  of  triangle. 

— =18  =  length  of  base  of  triangles. 

Then,  18,  +  262  =  1000,  and  V 1000 =31. 6223 = chord  of  arc  of  segments 

a  b,  c  d. 

To  ascertain  the  Versed  Sine.  From  the  square  of  the  radius  subtract 
the  square  of  half  the  chord,  and  the  square  root  of  the  remainder  sub- 
tracted from  the  radius  is  the  versed  sine. 

Thus,         100-^-2=50,  and  502= 2500= square  of  radius. 

31.0228-^2  =  15.8114,  and  15.81142=250=sgware  of  half  the  chord. 
2500-250=2250,  and  V 2250=47.4342 =square  root  of  the  difference 

of  the  squares  of  the  radius  and  half  the  chord. 
Then,  50-47.4342  =  2. 5658  =  versed  sine. 

Having  obtained  the  chord  of  the  arcs  (31.6228),  their 
versed  sines  (2.5658),  and  the  diameter  of  the  circle  (100), 
then,  by  rule,  page  75, 

VlOOx  2.5658  =  16.0181  =chord  of  half  the  arc. 

And  by  rule,  page  78,  to  ascertain  the  length  of  an  arc, 


MENSURATION    OF   AREAS,  LINES,  AND    SURFACES.  97 


16.0181  x  2  x  10  x  2.5658=821.9848=ta;ice  the  chord  of  half 

the  arc  by  10  times  the  versed  sine. 
100  x  60  —  2.5658  x  27=5830.7234  =  27  times  the  versed  sine 
subtracted  from  60  times  the  diameter. 

821.984:8+5930.7234:  =  .l385=quotientofaboveproductand 
remainder,  and  .1385  +  32.0362  (16.0181  x  2)  =32.1747 
=  length  of  the  arc. 
32. 1747x50 Tl  —  804.3675  =the  product  of  the  length  of  the 
arc  and  half  the  radius  of  the  circle =area  of  sector. 

And  804.3675-31-6228^47-4342=54.3664=amz  ofthetri- 

angle  subtracted  from  the  area  of  the  sector— area  of  each  seg- 
ment. 

54.3564x2         =   108.7 328=  area  of  segments. 

Area  of  trapezoid  =  2028. 

2136.7328  —area  of  zone. 

Ex.  2.  The  greater  chord  is  24,  the  lesser  15,  and  the 
breadth  of  the  zone  6.5  inches ;  what  is  its  area  ? 

(Length  of  arc  8.0437.)  Ans.  133.5467  inches. 

Ex.  3.  The  lesser  chord  is  96,  the  greater  100,  and  the 
breadth  of  the  zone  14  inches  ;  what  is  the  area  of  it  % 

(Length  of  arc  14.189.)  Ans.  1381.4851  inches. 

Ex.  4.  The  chords  of  a  zone  are  96  inches,  and  its  height 
28  ;  what  is  its  area  1 

(Length  of  arc  28.379.)  Ans.  2762.95  inches. 

Centre  of  Gravity.  Find  the  centres  of  gravity  of  the  trap- 
ezoid and  the  segments  comprising  the  zone;  draw  a  line 
(equally  dividing  the  zone)  perpendicular  to  the  chords ;  con- 
nect the  two  centres  of  the  segments  by  a  line  cutting  the  per- 
pendicular to  the  chords ;  then  will  the  centre  of  gravity  of 
the  figure  be  on  the  perpendicular,  toward  the  lesser  chord, 
at  such  proportionate  distance  of  the  difference  between  the 
centres  of  gravity  of  the  trapezoid  and  line  connecting  the  cen- 
tres of  the  segments  as  the  area  of  the  two  segments  bears  to 
the  area  of  the  trapezoid. 

E 


98 


MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 


Cylinder. 

Definition.  A  figure  formed  by  the  revolution  of  a  right- 
angled  parallelogram  around  one  of  its  sides. 

To  ascertain  tlie  Surface  of  a  Cylinder  {Fig.  37). 

Eule. — Multiply  the  length,  a  b,  by  the  circumference,  and 
the  product  added  to  the  area  of  the  two  ends  will  be  the  sur- 
face required. 

Or,  lxc-\-2  a=zs,  where  a  represents  area  of  end. 

Note. — "When  the  internal  surface  alone  is  wanted,  the  areas  of  the 
ends  are  to  be  omitted. 

Fig,  37. 


Example. — The  diameter  of  a  cylinder,  b  c,  is  30  inches, 
and  its  length,  a  b,  50  inches  ;  what  is  its  surface? 
30  x  3.1416  =  94.2480  inches = circumference. 

94.248  x50=4712A=area  of  body. 

And  302x- 7854  =  706.86  =  area  of  one  end. 

706.86  X  2  =  1413.72 =amz  of  both  ends. 

Then,  4712.4  +  1413.72  =  6126.12=swr/ace  required. 

Ex.  2.  The  diameter  of  a  cylinder  is  100  inches,  and  its 

length  12  feet;  what  is  its  surface*?         Ans.  423.243  feet. 

Ex.  3.  The  diameter  of  a  hollow  cylinder  is  36  inches,  and 
its  length  10  feet;  what  is  its  internal  surface? 

Ans.  94.248  feet. 
Centre  of  Gravity.     Is  in  its  geometrical  centre. 


MENSURATION    OF    AREAS,  LINES,  AND    SURFACES. 


99 


Prisms. 

Definition.     Figures  the  sides  of  which  are  parallelograms,  and 

the  ends  equal  and  parallel. 

Note. — When  the  ends  are  triangles,  they  are  called  triangular 
prisms;  when  they  are  square,  they  are  called  square  or  right  prisms ; 
when  they  are  pentagons,  pentagonal  prisms,  &c,  &c. 

To  ascertain  the  Surface  of  a  Prism  (Figs.  38  and  39). 

Kule. — Find  the  areas  of  the  ends  and  sides  as  by  the  rules 
for  the  mensuration  of  squares,  triangles,  &c.,  and  add  them 
together ;  the  sum  will  be  the  surface  of  the  figure. 

Or,  2  a-{-a/—s,  where  a  represents  the  area  of  the  ends,  and 
a/  the  area  of  the  sides. 

Fig  38.  a  b  Fig.  39.      b 


Example. — The  side  a  b,  Fig.  38,  of  a  square  prism  is  12 
inches,  and  the  length,  b  c,  30;  what  is  the  surface  % 
12  x  12  =  144=area  of  one  end. 
144  x  2  =  288  —area  of  both  ends. 
12  x  30  —  360 —area  of  one  side. 
360x4=1440=araz  of  four  sydes. 
Then,  2884-1440  =  1728  inches,  the  surface  required. 
Ex.  2.  What  is  the  surface  of  a  triangular  prism,  the  sides 
ab,bc,  and  c  a,  Fig.  39,  being  each  12  inches,  and  the  length, 
c  d,  30  inches'? 

12-^2  =  6,  and  VQ2  —  122= 10. 3923= width  of  prism. 
Hence,  10.3923  x  1 2 -j- 2  =  62.3538^=  area  of  each  end. 


100        MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

12  x  30  =  360= area  of  one  side. 
62.3538  x  2  =  124.7076 —area  of  ends, 
Then,  360  x  3  =  1080 = area  of  sides, 

and  124.7076  +  1080=1204.7076  inches— surface  required. 

Ex.  3.  What  is  the  surface  of  a  rhomboidal  prism,  the  depth 
of  it  being  5  feet  9  inches,  the  width  7  feet,  and  the  length  10 
feet?  Ans.  A02.5  feet. 

Centre  of  Gravity.  When  the  ends  are  parallelograms,  it  is 
in  their  geometrical  centre. 

When  the  ends  are  triangles,  trapeziums,  etc.,  it  is  in  the  mid- 
dle of  their  length  at  the  same  distance  from  the  base  as  that 
of  the  triangle  or  trapezoid  which  is  a  section  of  them. 

Wedge. 

Definition.     A  wedge  is  a  prolate  triangular  prism,  and  its 
surface  is  found  by  the  rule  for  that  of  a  right  prism. 
Fig.  40.        e 


b  c 

Example. — The  back  of  a  wedge,  abed,  Fig.  40,  is  20  by 

2  inches,  and  its  end,  ef,  20  by  2  inches ;  what  is  its  surface? 

2 

202+2-f- 1  —401  =sum  of  the  squares  of  half  the  base,  a  f,  and 

the  height,  ef,  of  the  triangle,  efa. 
y^401  =  20.025  —square  root  of  above  sum  =  length  of  e  a. 
Then,  20.025  x  20  x  2=801  —area  of  sides. 
And  20 x 2=4,0— area  of  bach,  and  20x2-4-2  x 2=40=area 

of  ends. 
Hence,  8014-40  +  40  =  881=  surface  required. 

Centre  of  Gravity.     See  rule  for  prisms. 


MENSURATION   OF  AREAS,  LINES,  AND  SURFACES.         101 


Prismoids. 

Definition.     Figures  alike  to  a  prism,  but  having  only  one  pair 
of  thdr  sides  parallel. 

To  ascertain  the  Surface  of  a  Prismoid  {Fig.  41). 

Rule. — Find  the  area  of  the  ends  and  sides  as  by  the  rules 
for  squares,  triangles,  &c.,  and  add  them  together. 

Fig.  41.     a  b 


g  h 

Example. — The  ends  of  a  prismoid,  efgh  and  abed,  Fig. 
41,  are  10  and  8  inches  square,  and  its  slant  height  25  ;  what 
is  its  surface  ? 

10  x  10  =  100=arai  of  base. 
8  x    8  =   64 = area  of  top. 

— i-x25  =  225,  and  225  x4=1000=ara*  of  sides. 

It 

Then,        100  +  64  + 1000= \\^— surface  required. 

Ex.  2.  The  ends  of  a  prismoid  are  15  and  12  inches  square, 
and  its  slant  height  40  ;  what  is  its  surface  % 

Ans.  2529  inches.    . 
Ex.  3.  The  ends  of  a  prismoid  are  12x16  and  14x18  inches, 
and  its  vertical  height  is  33  ;  what  is  its  surface  % 

Ans.  2424  inches. 

Centre  of  Gravity.  Is  at  the  same  distance  from  its  base 
as  that  of  the  trapezoid  or  trapezium  which  is  a  section  of  it. 


102        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 


Ungulas. 

Definition.  Cylindrical  ungulas  are  frustrums  of  cylinders. 
Conical  ungulas*  are  frustrums  of  cones. 

To  ascertain  the   Curved  Surface  of  an  Ungula,  Figs.  42,  43, 
44,  45,  and  46. 

1.  When  the  Section  is  parallel  to  the  Axis  of  the  Cylinder,  Fig.  42. 

Rule. — Multiply  the  length,  a  b  c,  of  the  arc  line  of  one 
end  by  the  height,  b  d,  and  the  product  will  be  the  curved 
surface  required. 

Or,  cxh=zs,  where  c  represents  length  of  arc  line. 
Fig.  42.         a 


Example. — The  diameter  of  a  cylinder  from  which  an  un- 
gula is  cut  is  10  inches,  its  length  50,  and  the  versed  sine  or 
depth  of  the  ungula  is  5  inches ;  what  is  the  curved  surface 
of  it? 

10-4-2=5  =  radius  of  cylinder. 
Hence  the  radius  and  versed  sine  are  equal;  the  arc  line,  therefore, 

of  the  ungula  is  one  half  the  circumference  of  the  cylinder, 

which  w  31.416-^2  =  15.708,       v 
and  15.708x50=785.400  inches.  • 

Ex.  2.  The  base  line  of  the  section  of  a  cylindrical  ungula 
is  48  inches,  the  height  or  versed  sine  of  the  arc  is  20,  and 

*  For  mensuration  of  conical  ungulas,  see  Conic  Sections,  p.  253. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


103 


the  length  of  the  ungula  is  20.5  feet ;  what  is  its  curved  sur- 
face? Ans.  109.8388  feet. 

2.  When  the  Section  passes  obliquely  through  the  opposite  Sides  of 
the  Cylinder,  Fig.  43. 

Rule. — Multiply  the  circumference  of  the  base  of  the  cylin- 
der by  half  the  sum  of  the  greatest  and  least  heights,  d  b  and 
e  a,  of  the  ungula,  and  the  product  will  give  the  curved  surface 
required. 

Fig.  43. 


Example. — The  diameter  of  a  cylindrical  ungula  is  10 
inches,  and  the  greater  and  less  heights  are  25  and  15  inches ; 
what  is  its  surface  ? 

10  diameter = 31.416  circumference. 
25  +  15=40,  and  40-=- 2  =  20. 
Hence,  31.416  x  20=62.8320  inches. 

Ex.  2.  The  circumference  of  an  ungula  is  60.75  inches,  and 
the  mean  height  of  it  13  feet ;  what  is  its  surface? 

Ans.  65.8125  feet. 

3.  When  the  Section  passes  through  the  Base  of  the  Cylinder  and  one 

of  its  Sides,  and  the  Versed  Sine  does  not  exceed  the  Sine,  Fig.  44. 

Rule. — Multiply  the  sine,  a  d,  of  half  the  arc,  d  g,  of  the 

base,  d  g  f,  by  the  diameter,  e  g,  of  the  cylinder,  and  from  this 

product  subtract  the  product*  of  the  arc  and  cosine,  a  o.    Mul- 

*  When  the  cosine  is  0,  this  product  is  0. 


104        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

tiply  the  difference  thus  found  by  the  quotient  of  the  height, 
g  c,  divided  by  the  versed  sine,  a  g,  and  the  product  will  be 
the  curved  surface  required. 

Fig.  44. 


Example. — The  sine,  a  d,  of  half  the  arc  of  the  base  of  an 
ungula  is  5,  the  diameter  of  the  cylinder  is  10,  and  the  height 
of  the  ungula  10  inches ;  what  is  the  curved  surface  ? 
5  x  10  =  50= sine  of  half  the  arc  hy  the  diameter. 
Length  of  arc,  the  versed  sine  and  radius  being  equal,  under  ride, 

page  78  =  15.708. 
Again,  as  the  versed  sine  and  the  radius  are  equal,  the  cosine  is  0. 
Hence,  when  the  cosine  is  0,  the  product  is  0.     50  —  0  =  50  =  the 

difference  of  the  product  before  obtained  and  the  product  of  t/ie 

arc  and  the  cosine. 
50xlO-f-5  =  50x2  =  100=*Ae  difference  multiplied  by  the  height 

divided  by  the  versed  sine,  which  is  the  surface  required. 

Ex.  2.  The  sine  of  half  the  arc  of  the  base  of  an  ungula  is  12 
inches,  the  versed  sine  is  9,  the  diameter  of  the  cylinder  is  25, 
and  the  height  of  the  ungula  is  18  ;  what  is  its  curved  surface l? 

12  x  25  =  300  =product  of  sine  and  diameter. 
Arc  of  base  of  ungula,  by  rule,  p.  77,  the  versed  sine  being  9,  is 

32*14795. 
Then,  32.14795  x  12.5-9  =  112.51782, 

and  300-112.51782  =  187.48218,-  which,  multiplied  by  18-f-9 

=  374,96436  inches. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


105 


4.  When  the  Section  passes  through  the  Base  of  the  Cylinder,  and 
the  Versed  Sine  exceeds  the  Sine,  a  g,  Fig.  45. 

Rule. — Multiply  the  sine  of  half  the  arc  of  the  base  by  the 
diameter  of  the  cylinder,  and  to  this  product  add  the  product 
of  the  arc  and  the  excess  of  the  versed  sine  over  the  sine  of 
the  base. 

Multiply  the  sum  thus  found  by  the  quotient  of  the  height 
divided  by  the  versed  sine,  and  the  product  will  be  the  curved 
surface  required. 

Fig.  45. 

/"" "\ 

i  1 


Example. — The  sine,  a  d,  of  half  the  arc  of  an  ungula  is 
12  inches,  the  versed  sine,  a  g,  is  16,  the  height,  eg,  16,  and 
the  diameter  of  the  cylinder  25  inches ;  what  is  the  curved 

surface  ? 

12  x2o=30Q=sine  of  half  the  arc  by  the  diameter  of  the  cyl- 
inder. 

Length  of  arc  of  base,  by  rule,  p.  74:=a?v  of  d  b  f— circumfer- 
ence of  base— 46.392. 

Then  46.392  x  16  — 12.5  =  162.372  ^product  of  arc  and  the  ex- 
cess of  the  versed  sine  over  the  sine. 

300-|-162.372=462.372  =  *Ae  sum  of  the  above  products. 
16  -i-16  =  l= quotient  of  height  divided  by  the  versed  sine. 

462.372x1  =  462.372  inches=the  sum  of  the  products  and  the 

height  divided  by  the  versed  sine  =  the  curved  surface  required. 

E2 


106        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 


Ex.  2.  The  sine  of  half  the  arc  of  the  base  of  an  ungula  is 

0,  the  diameter  of  the  cylinder- is  10  inches,  and  the  height  of 

the  ungula  is  20  inches;  what  is  its  curved  surface? 

Note. — The  sine  of  the  arc  being  0,  the  versed  sine  is  equal  to  the 
diameter  (10),  and  the  sine  of  the  base  is  10-^2=5. 

0x  10  =  0  = product  of  sine  of  half  the  arc  and  diameter  of  the 
cylinder. 

0-f(31.416  {length  of  arc)  x  10oo5)=  157.08  —  the  sum  of  the 
product  above  obtained  and  the  product  of  the  arc  and  the  ex- 
cess of  the  versed  sine  over  the  sine. 

157.08  x 20-r-10=314.16  =  ^e   above  sumxthe   height-i-the 
versed  sine=the  result  required. 

5.  When  the  Section  passes  obliquely  through  both  Ends  of  the 
Cylinder,  abed.  Fig.  46. 

Rule. — Conceive  the  section  to  be  continued  till  it  meets 
the  side  of  the  cylinder  produced ;  then,  as  the  difference  of 
the  versed  sines  of  the  arcs  of  the  two  ends  of  the  ungula  is 
to  the  versed  sine  of  the  arc  of  the  less  end,  so  is  the  height 
of  the  cylinder  to  the  part  of  the  side  produced. 

Find  the  surface  of  each  of  the  ungulas  thus  found  by  the 
rules  3  and  4,  and  their  difference  will  be  the  curved  surface 
required. 

Fig.M.      k 


Example. — The  versed  sines  a  e,  d  o,  and  sines  i  Tc,  g  r,  of 


MENSURATION    OP  AREAS,  LINES,  AND    SURFACES.         107 

the  arcs  of  the  two  ends  of  an  ungula,  Fig.  46,  are  respectively 
5  and  2.5,  and  5.and  4.25  ;  the  height  of  the  ungula  within  the 
cylinder,  cut  from  one  having  10  inches  diameter,  is  5  inches  ; 
what  is  the  height  of  the  ungula  produced  beyond  the  cylinder? 
5cv>2.5  =  2.5,  and  2.5  :  2.5  ::  5  :  5= height  of  ungula  produced 

beyond  the  cylinder, 

Ex.  2.  The  versed  sines  of  the  base  and  arc  of  an  ungula 
cut  from  a  cylinder  of  6  inches  diameter  are  6  and  2  inches, 
and  its  height  within  the  cylinder  is  4  inches ;  what  is  the 
distance  it  extends  above  or  beyond  the  cylinder  ! 

Ans.  2  inches. 

Lune. 

Definition.  The  space  between  the  intersecting"  arcs  of  two 
eccentric  circles. 

To  ascertain  the  Area  of  a  Lune,  Fig.  47. 

Kule. — Find  the  areas  of  the  two  segments  from  which  the 

lune  is  formed,  and  their  difference  will  be  the  area  required. 

Or,  s—s'=a,  wlien  s  and  s/  represents  the  areas  of  the  segments. 

Fig.  47.  d 


Example. — The  length  of  the  chord  a  c  is  20,  the  height 
e  d  is  3,  and  e  b  2  inches ;  what  is  the  area  of  the  lune  ? 
By  Rule  2,  p.  76,  the  diameters  of  the  circles  of  which  the  lune  is 

formed  are  thus  found: 

lor  ad  c,         ^ —=25. 

5 

102+22     ffo 
For  a  e  c,  - —  52. 


108        MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 

Then,  by  rule  for  the  areas  of  segments  of  a  circle,  page  87, 
the  area  of  a  d  c  is  70.5577  in. 
.    "         a  e  c      27.1638  in. 
their  difference  43.3939  in.,  the  area  of  the  lune  required. 

Ex.  2.  The  chord  of  a  lune  is  40,  and  the  heights  of  the 
segments  10  and  4  inches;  what  is  its  content? 

Ans.  173.5752  inches. 
'  Ex.  3.  The  chord  of  a  lune  is  6  feet  8  inches,  and  the 
heights  of  the  arcs  1.666  feet  and  8  inches;  what  is  its  area? 

Ans.  694.2996  inches. 
Ex.  4.  The  chord  of  a  lune  is   86.6024  inches,  and  the 
heights  of  the  segments  25  and  15  inches;  what  is  its  area? 

Ans.  653.3551. 

Centre  of  Gravity.  On  a  line  connecting  the  centres  of 
gravity  of  the  two  arcs  at  a  point  proportionate  to  the  respect- 
ive areas  of  the  arcs. 

Note. — If  semicircles  be  described  on  the  three  sides  of  a  right- 
angled  triangle  as  diameters,  two  lunes  will  be  formed,  and  their  united 
areas  will  be  equal  to  that  of  the  triangle. 

Cycloid. 

Definition.  A  curve  generated  by  the  revolution  of  a  circle  on 
a  plane. 

To  ascertain  the  Area  of  a  Cycloid,  Fig.  48. 

Rule. — Multiply  the  area  of  the  generating  circle  a  b  c  by 
3,  and  the  product  will  give  the  area  required. 
Or,  a x^— area. 

Fig.  48. 


MENSURATION    OP  AREAS,  LINES,  AND    SURFACES.         109 

Example. — The  generating  circle  of  a  cycloid  has  an  area 
of  115.45  inches  ;  what  is  the  area  of  the  cycloid1? 
115.45  x  3  =  346.35  inches. 
Ex.  2.  The  area  of  a  circle  describing  a  cycloid  is  1.625 
feet;  what  is  the  area  of  the  cycloid  in  inches? 

Ans.  702  inches. 
Ex.  3.  The  diameter  of  a  circle  describing  a  cycloid  is 
66.5  feet ;  what  is  the  area  of  the  cyloid  in  inches  ? 

Ans.  1500434.064  inches. 

To  ascertain  the  Length  of  a  Cycloidal  Curve,  Fig.  48. 

Rule. — Multiply  the  diameter  of  the  generating  circle  by 
4,  and  the  product  will  give  the  length  of  the  curve. 

Or,  d  x  4  =  length  of  curve. 

Example. — The  diameter  of  the  generating  circle  of  a  cy- 
cloid, Fig.  48,  is  8  inches ;  what  is  the  length  of  the  curve  dsc? 

8  x  4= 32 '= product  of  diameter  and  4  =  the  length  required. 

Ex.  2.  The  diameter  of  the  generating  circle  is  20  inches  ; 
what  is  the  length  of  the  cycloidal  curve  % 

Ans.  80  inches. 

Centre  of  Gravity.  At  a  distance  from  the  centre,  n,  of 
the  chord,  d  c,  of  the  curve  d  s  c— -|  of  the  radius  of  the  gen- 
erating circle. 

Note. — The  curve  of  a  cycloid  is  the  line  of  swiftest  descent ;  that  is, 
a  body  will  fall  through  the  arc  of  this  curve,  from  one  point  to  another, 
in  less  time  than  through  any  other  path. 

RINGS. 

Circular  Rings. 
Definition.     The  space  betiveen  two  concentric  circles. 

To  ascertain  the  Sectional  Area  of  a  Circular  Ring,  Fig.  49. 

Rule. — From  the  area  of  the  greater  circle,  a  b,  subtract 
that  of  the  less,  c  d,  and  the  difference  will  be  the  area  of  the 
ring. 

Or,  a— a'  =:  area. 


110       MENSURATION   OF   AREAS,  LINES,  AND    SURFACES. 
Fig.  49. 


Example. — The  diameters  of  the  circles  forming  a  ring 
are  each  10  and.,15  inches  ;  what  is  the  area  of  the  ring? 
Area  of  15  =  176.7146 
"        10=  78.5400 

98.1746  inches. 
Ex.  2.  The  diameters  of  a  circular  ring  are  10.75  and  18.25 
inches;  what  is  its  area?  Ans.  171.82  inches. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 


Definition. 


Cylindrical  Rings. 
A  ring  formed  by  the  curvature  of  a  cylinder. 


To  ascertain  the  Convex  Surface  of  a  Cylindrical  Hing,  Fig.  50. 

Kule. — To  the  thickness  of  the  ring,  a  b,  add  the  inner 
diameter,  be;  multiply  this  sum  by  the  thickness  and  the 
product,  by  9.8696,  and  it  will  give  the  surface  required. 

Or,  d+d' xdx9.S696=surface. 
Fig.  50. 


Example. — The  thickness  of  a  cylindrical  ring,  a  b,  is  2 
inches,  and  the  inner  diameter,  b  c,  is  18 ;  what  is  the  surface 
of  it? 


MENSURATION   OF  AREAS,  LINES,  AND  SURFACES. 


Ill 


2  +  18  —  20  =  thickness  of  ring  added  to  the  inner  diameter. 
20X  2  X  9. 8696  =  394.784  =  Z/*e  sum  above  obtained X  the  thick- 
ness of  the  ring,  and  that  product  by  9.8696,  the  result  re- 
quired. 

Ex.  2.  The  thickness  of  a  ring  of  metal  of  20  inches  diam- 
eter (internal)  is  2  inches  ;  what  is  the  surface  of  it  1 

Ans.  434.2624  inches. 


Link. 
Definition.     An  elongated  ring. 

To  ascertain  the  Convex  Surface  of  a  Link,  Figures  51. 

Rule. — Multiply  the  circumference  of  a  section  of  the  body, 
a  b,  of  the  link  by  the  length  of  its.  axis,  and  the  product  .will 
give  the  surface  required. 

Or,  c  x  l=swface. 

Note. — To  ascertain  the  Circumference  or  Length  of  the  Axes. 

When  the  Ring  is  elongated.  To  the  less  diameter  add  its  thickness, 
multiply  the  sum  by  3.1416  ;  multiply  the  difference  of  the  diameters  by 
2,  and  the  sum  of  these  products  will  give  the  result  required. 

When  the  Ring  is  elliptical.  Square  the  diameters  of  the  axes  of  the 
ring,  and  multiply  the  square  root  of  half  their  sum  by  3.141G;  the 
product  will  give  the  length  of  the  body  of  the  ring. 

Figs.  51. 


Example. — The  link  of  a  chain  is  1  inch  in  diameter  of 
body,  a  b,  and  its  inner  diameters,  b  c  and  ef,  are  12.5  and  2.5 
inches ;  what  is  its  circumference. 
2.5  +  1x3.1416  —  10.995Q  =  length  of  axis  of  ends. 
12.5  —  2.5x2x2  =  15  =  length  of  sides  of  body. 
Then,  10.9956  + 15  =  25.9956  =  length  of  axis  of  link,  which, 

X  3.1416  (cir.  of  1  in.) ^  1.6678  inches. 

Centres  of  Gravity.     Are  in  their  geometrical  centres. 


112        MENSURATION    OP  AREAS,  LINES,  AND    SURFACES. 


Cones. 

Definition.  A  figure  described  by  the  revolution  of  a  right- 
angled  triangle  about  one  of  its  legs. 

For  Sections  of  a  Cone,  see  Conic  Sections,  page  228. 

To  ascertain  the  Surface  of  a  Cone,  Fig.  52. 

Kule. — Multiply  the  perimeter,  or  circumference  of  the 
base,  by  the  slant  height,  or  side  of  the  cone,  and  half  the 
product  added  to  the  area  of  the  base  will  be  the  surface. 

Or,  — — -  —surface. 

Fig.  52.        „ 


Example. — The  diameter,  a  b,  of  the  base  of  a  cone  is  3  feet, 
and  the  slant  height,  a  c,  15  ;  what  is  the  surface  of  the  cone  ? 

Perimeter  of  3  fcet= 9.4248,  and  9'424^  X  15^70.686- 

m 

surface  of  side  of  cone. 
Area  of  3  feet=7. 068,  and  70.686 +7.068  =  77.754 ^surface 
required. 

Ex.  2.  The  diameter  of  the  base  of  a  cone  is  6.25  inches, 
and  the  slant  height  18.75  ;  what  is  the  surface  of  it? 

Ans.  214.757  inches. 
Ex.  3.  The  diameter  of  the  base  of  a  cone  is  20  inches,  and 
the  slant  height  14.142  ;  what  is  the  surface  of  the  cone? 

Ans.  758.445  inches. 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.        113 

To  ascertain  the  Surface  of  the  Frustrum  of  a  Cone,  Fig.  53. 

Rule. — Multiply  the  sura  of  the  perimeters  of. the  two  ends 
by  the  slant  height  of  the  frustrum,  and  half  the  product  add- 
ed to  the  areas  of  the  two  ends  will  be  the  surface  required. 

_     p+p'xh  .      .    ,  , 

Or,  - — ~ \-a-\-a  —  surface. 

& 

Fig.  53. 


Example. — The  frustrum,  abed,  Fig.  53,  has  a  slant 
height  of  26  inches,  and  the  circumferences  of  its  ends  are 
15.7  and  22.  inches  respectively;  what  is  its  surface? 
15.7  +  22.  x  26^-2— 490. 1  =su?face  of  sides. 

(i*7\2  /22\2 

HiTe)  *-7854+(04l6)  X  .7854=58.12=™/^, 
Then,  490.1  +  58.12  =  548.22  ^surface. 

Ex.  2.  What  is  the  surface  of  the  frustrum  of  a  cone,  the 
diameters,  of  the  ends,  a  c  and  b  d,  being  4  and  8  feet,  and  the 
length  of  the  slant  sides  20  feet !  Ans.  439.824  feet. 

Ex.  3.  What  is  the  surface  of  the  frustrum  of  a  cone,  the 
diameter  of  the  ends  being  6.66  and  10  feet,  and  the  length 
of  the  slant  side  3.73  feet  ? 

Ans.  210.989. 

Centres  of  Gravity.  Cone  or  Frustrum. — At  the  same  dis- 
tance from  the  base  as  in  that  of  the  triangle  or  parallelo- 
gram, which  is  a  right  section  of  them. 


114        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 


Pyramids. 

Definition.  A  figure,  the  base  of  which  has  three  or  more 
sides,  and  the  sides  of  which  are  plane  triangles. 

To  ascertain  the  Surface  of  a  Pyramid,  Fig.  54. 

Eule. — Multiply  the  perimeter  of  the  base  by  the  slant 
height,  and  half  the  product  added  to  the  area  of  the  base 
will  be  the  surface. 

r\     Pxh  i  * 

Or,  — - — \- a— surface. 


Fig.  55. 


be  b 

Example. — The  side  of  a  quadrangular  pyramid,  a  b,  Fig.  54, 
is  12  inches,  and  its  slant  height,  c  e,  40 ;  what  is  its  surface? 


12x4=48 
48x40 


perimeter  of  base. 
960  ="  area  of  sides. 


Then  12  X  12  +  960  =  1104=swr/ace. 

Ex.  2.  The  sides  of  a  hexagonal  pyramid  are  12.5  inches, 
and  its  slant  height  62.5  ;  what  is  its  surface? 

Ans.  2749.703  inches. 

Ex.  3.  The  sides,  a  b,  b  c,  of  an  oblong  quadrangular  pyr- 
amid, Fig.  55,  are  15  and  17.5  inches,  and  its  slant  height, 
d  e,  36  ;  what  is  its  surface?  Ans.  1432.5  inches. 

Ex.  4.  The  sides  of  an  octagonal  pyramid  are  4  feet  2 
inches,  and  its  slant  height  6  feet  9  inches ;  what  is  its  sur- 
face? Ans.  196.326  square  feet. 


MENSURATION   OF  AREAS,  LINES,  AND    SURFACES.        115 

Ex.  5.  What  is  the  surface  of  a  pentagonal  pyramid,  its 
slant  height  being  12  feet,  and  each  side  of  its  base  2  feet? 

Ans.  66.882. 

To  ascertain  the  Surface  of  the  Frustrum  of  a  Pyramid,  Fig.  56. 

Rule. — Multiply  the  sum  of  the  perimeters  of  the  two  ends 

by  the  slant  height  or  side,  and  half  the  product  added  to  the 

area  of  the  ends  will  be  the  surface. 

_     p+p'xh  ,  „ 

Or,  - — ^r \-a+a  =  surface. 

Fig.  56.         d  c 


ExAarPLE. — The  sides,  a  b,  c  d,  Fig.  56,  of  a  quadrangular 
pyramid  are  10  and  9  inches,  and  its  slant  height,  e  o,  20 ; 
what  is  its  surface  1 

10x4=40 
9x4=36 

7Q=sum  of  perimeters. 

76X20  =  1520,  and  -—— =7 60= area  of  sides. 
z 

10x10=100,  and  9x9=81. 
Then        100  +  81+760  =  941,  the  surface. 

Ex.  2.  The  ends  of  a  frustrum  of  a  quadrangular  pyrami 
are  15  and  9  inches,  and  its  slant  height  40 ;  what  is  its  sur- 
face? .  Ans.  2226  inches. 

Ex.  3.  The  ends  of  a  frustrum  of  a  triangular  pyramid  are 
20  and  10  inches,  and  its  slant  height  50 ;  what  is  its  surface  ? 

Ans.  2466.5  inches. 

Ex.  4.  The  sides  of  a  frustrum  of  a  hexagonal  pyramid  are 
15  and  25  inches,  and  its  slant  height  20 ;  what  is  its  sur- 
face? .  Ans.  32.002  square  feet. 


116         MENSURATION    OF   AREAS,  LINES,  AND    SURFACES. 

Centres  of  Gravity.  Pyramid  or  Frustrum. — At  the  same 
distance  from  the  base  as  in  that  of  the  triangle  or  parallelo- 
gram, which  is  a  right  section  of  them.. 

Helix  {Screw). 

Definition.  A  line  generated  by  the  progressive  rotation  of  a 
point  around  an  axis  and  equidistant  from  its  centre. 

To  ascertain  the  Length  of  a  Helix,  Fig.  57. 

Rule. — To  the  square  of  the  circumference  described  by 
the  generating  point,  add  the  square  of  the  distance  advanced 
in  one  revolution,  and  the  square  root  of  their  sum  multiplied 
by  the  number  of  revolutions  of  the  generating  point  will  give 
the  length  of  the  line  required. 

Or,  ^{c2-\-h2)xn— length  of  line,  n  representing  the  number 
of  revolutions. 

Fig.  57. 


Example. — What  is  the  length  of  a  helical  line  running 

3.5  times  around  a  cylinder  of  22  inches  in  circumference  and 

advancing  16  inches  in  each  revolution1? 

222  + 162  =  740 =sum  of  squares  of  circumference  and  of  the  dis- 
tance advanced. 

-y/740  X  3.5  =  95.21  —square  root  of  above  sum  x  number  of  rev- 
olutions ==  length  of  line  required. 
Ex.  2.  What  is  the  length  of  the  helical  line  described  by 

a  point  in  a  screw  in  one  revolution  at  a  radius  from  its  axis 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.        117 

of  11.3  inches,  the  progression  of  the  line  or  pitch  of  the 
screw  being  17  inches  ?  Ans.  73. 

Ex.  3.  What  is  the  length  of  the  helical  line  described  by 
a  point  on  the  periphery  of  a  screw  of  10  feet  in  diameter, 
having  a  pitch  of  20  feet?  Ans.  37.242. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

Spirals. 

Definition.  -Lines  generated  by  the  progressive  rotation  of  a 
point  around  a  fixed  axis. 

A  Plane  Spiral  is  when  the  point  rotates  around  a  central 
point. 

A  Conical  Spiral  is  when  the  point  rotates  around  an  axis 
or  a  cone. 

To  ascertain  the  Length  of  a  Plane  Spiral  Line,  Fig.  58. 

Rule. — Add  together  the  greater  and  less  diameters,*  di- 
vide their  sum  by  two,  multiply  the  quotient  by  3.1416,  again 
by  the  number  of  revolutions,  and  the  product  will  give  the 
length  of  the  line  required. 

Or,. when  the  circumferences  are  given,  take  their  mean 
length,  multiply  it  by  the  number  of  revolutions,  and  the  prod- 
uct will  give  the  length  required. 

Or, — — -  X  3.1416  Xn= length   of  line,  n   representing   the 

number  of  revolutions. 

Fig.  58. 


*  When  the  spiral  is  other  than  a  line,  measure  the  diameters  of  it 
from  the  middle  of  the  material  composing  it. 


118        MENSURATION    OP  AREAS,  LINES,  AND    SURFACES. 

Example. — The  less  and  greater  diameters  of  a  plane  spiral 
spring,  as  a  b,  c  d,  Fig.  58,  are  2  and  20  inches,  and  the  num- 
ber of  revolutions  10 ;  what  is  the  length  of  it  ? 

2  +  20 

— - — =11— swm  of  diameters -f-  2. 
z 

llX3.U16  =  34:.5576  =  above  quotient X 3.1416. 
34.5576  X  10  =  345.576  =above  product  X  number    of  revolu- 
tions z^the  length  of  line. 

Ex.  2.  The  greater  and  less  diameters  of  a  plane  spiral  are 
4  and  30  inches,  and  the  number  of  revolutions  5  ;  what  is 
the  length  of  it  %  Ans.  267.036  inches. 

To  ascertain  the  Length  of  a  Conical  Spiral,  Fig.  59. 

Rule.-1- Add  together  the  greater  and  less  diameters  ;*  di- 
vide their  sum  by  two,  and  multiply  its  quotient  by  3.1416. 

To  the  square  of  the  product  of  this  circumference  and  the 
number  of  revolutions  of  the  spiral,  add  the  square  of  the 
height  of  its  axis,  and  the  square  root  of  the  sum  will  be  the 
length  required. 

Or,  V\— g— X3.1416xrc  + A  =  length  of  spiral. 
Fig.  59.         c 


Example. — The  greater  and  less  diameters  of  a  conical 
spiral,  Fig.  59,  are  20  and  2  inches,  its  height,  c  d,  10,  and  the 
number  of  revolutions  10;  what  is  the  length  of  it? 

*  See  Note  to  Rule,  page  117. 


MENSURATION   OP  AREAS,  LINES,  AND    SURFACES.        119 

20+2_i-2  =  ll  x3.1416=34.5576=swm  of  diameters +2  and 

X3.1416. 
34.5576X10=345.576,  and    345.5762= 11 9422.77  =square 

of  the  product  of  the  circumference  and  number  of  revolutions. 
Vll9422.77  +  102=345.72  =  ^  square  root  of  the  sum  of  the 

above  product  and  the  square  of  the  height  of  the  spiral— the 

result  required. 

Ex.  2.  The  greater  and  less  diameters  of  a  conical  spiral  are 
1.5  and  8.75  feet,  its  height  6  feet,  and  the  number  of  its  revo- 
lutions is  5  ;  what  is  the  length  of  it  ?      Ans.  80.725  inches. 

Ex.  3.  The  greater  and  less  diameters  of  a  conical  spiral 
are  3  and  9  feet,  its  height  12.5  feet,  and  the  number  of  its 
revolutions  10;  what  is  the  length  of  it?         Ans.  188.91. 

Centres  of  Gravity.  Plane  Spiral. — It  is  in  its  geometrical 
centre. 

Conical  Spiral. — It  is  at  a  distance  from  the  base  J  of  the 
line  joining  the  vertex  and  centre  of  gravity  of  the  base. 

Note.^ — This  rule  is  applicable  to  winding  engines  where 
it  is  required  to  ascertain  the  length  of  a  rope,  its  thickness, 
the  number  of  revolutions,  diameter  of  drum,  etc.,  etc. 

Illustration. — The  diameter  over  the  roll  of  a  flat  rope  upon 
the  drum  of  a  winding  engine  shaft  is  134.5  inches,  the  diam- 
eter of  the  drum  is  94.5  inches,  and  the  number  of  revolutions 
20 ;  what  is  the  length  of  the  rope  and  what  is  its  thickness? 
Ans.  Length  of  the  rope,  7194.247  inches. 
Area  of  134.5  m.  =  14208.049 
"    "    94.5       =  7013.802 
7194.247 
Then,  7194.247 -^ 20  =  359.712 =area   of  rope~-revolutions= 

area  of  each  thickness.  # 

134.5-94.5-^2  +  94.5  =  114.5  and  114.5x3.1416  =  359.712 

= circumference  of  the  mean  diameter  of  each  thickness. 
Hence,  359.712 -f-359.7l2  =  l  inch—the  width  or  thickness  of 
the  rope. 

*  For  Rules  to  ascertain  the  elements  of  Winding  Engines,  see  Has- 
well's  Engineers'  and  Mechanics'  Pocket-book,  p.  263-4. 


120        MENSURATION   OF   AREAS,  LINES,  AND    SURFACES. 
.    SPINDLES. 

Definition.  Figures  generated  by  the  revolution  of  a  plane 
area,  ivhen  the  curve  is  revolved  about  a  chord  perpendicular  to 
its  axis,  or  about  its  double  ordinate,  and  they  are  designated  by 
the  name  of  the  arc  or  curve  from  which  they  are  generated,  as 
Circular,  Elliptic,  Parabolic,  etc.,  etc. 

Circular  Swindle, 

To  ascertain  the  Convex  Surface  of  a  Circular  Spindle,  Fig.  60. 

Rule. — Multiply  the  length,  /  c,  by  the  radius,  o  c,  of  the 
revolving  arc ;  multiply  this  arc,  fac,by  the  central  distance, 
o  e,  or  distance  between  the  centre  of  the  spindle  and  centre 
of  the  revolving  arc ;  subtract  this  product  from  the  former, 
double  the  remainder,  multiply  it  by  3.1416,  and  the  product 
will  be  the  surface  required. 

Or,  lxr—(aX  -y/r2  — I-)  )X2  p;  a  representing  the  length 

of  the  arc,  c  the  chord,  and  p  3.1416. 
Fig.  60.         a 


Example. — What  is  the  surface  of  a  circular  spindle,  Fig. 
60,  the  length  of  \t,fe,  being  14.142  inches,  the  radius  of  its 
arc,  o  c,  10,  and  the  central  distance,  o  e,  7.071  ? 
14.142x10  =  141.42  =  length  x  radius. 
Length  of  arc,  by  rules,  p.  76,  78  =  15.708. 
15.708  x  7.071  =  1 1 1.0713  —  length  of  arc  X  central  distance. 
141.42  —  111.0713=30.3487=^mice  of  products. 
30.3487  x  2  =  60.6974  x  3.1416  =  190.687  =  the  remainder 
doubled  x  3.1416,  which  is  the  result  required. 


MENSURATION   OF  AREAS,  LINES,  AND  SURFACES.         121 

Ex.  2.  The  length  of  a  circular  spindle  is  28.284  feet,  the 
radius  of  its  arc  20,  and  the  distance  between  the  centre  of 
the  spindle  and  the  centre  of  the  revolving  arc  is  14.142 ; 
what  is  the  surface  of  it  *  Ans.  762.7484  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Convex  Surface  of  a  Zone  of  a  Circular  Spin- 
dle, Fig.  61. 

Rule. — Multiply  the  length,  i  c,  by  the  radius,  o  «,  of  the 
revolving  arc  ;  multiply  the  arc,  d  a  b,  by  the  central  distance 
o  e;  subtract  this  product  from  the  former,  double  the  re- 
mainder, multiply  it  by  3.1416,  and  the  product  will  be  the 
surface  required. 

Or,  lxr—(ax  yV2— ( -J  )  x2p,  I  representing  t/ie  length  oj 
the  zone. 


•v/ 

O 


Example. — What  is  the  convex  surface  of  the  zone  of  a 
circular  spindle,  Fig.  61,  the  length  of  it  being  7.653  inches, 
the  radius  of  its  arc  10,  the  central  distance  7.071,  and  the 
length  of  its  side  or  arc,  d  b,  7.854  inches  ? 

7.653  X  10=76.53=lengthxradius. 

7.854:X7.071=55.5S56  =  length  of ' arcX central  distance. 

76.53— 55.5356= 20.9944 =^rerace  of  products. 

20.9944x2=41.9888x3.1416  =  131.912  =  ^6  remainder 

doubled  X  3.1416,  which  is  the  result  required. 
-   Ex.  2.  The  zone  of  a  circular  spindle  is  23  inches  in  length, 

F 


122        MENSURATION   OF  AREAS,  LINES,  AND   SURFACES. 

the  radius  of  its  arc  30,  its  central  distance  21.2,  and  the 
length  of  its  side  23.56  ;  What  is  its  convex  surface  ? 

Ans.  1197.2255  inches. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Convex  Surface  of  a  Segment  of  a  Circular 
Spindle,  Fig.  62. 

Kule. — Multiply  the  length,  i  c,  by  the  radius  of  the  re- 
volving arc,  o  a;  multiply  the  arc  by  the  central  distance,  o  e  ; 
subtract  this  product  from  the  former,  double  the  remainder, 
multiply  it  by  3.1416,  and  the  product  will  be  the  surface 
required. 


Or,  lxr—(ax  \/r2—  I  -  )  )  X  2p,  I  representing  the  length  of 
the  segment. 


o 
Example. — What  is  the  convex  surface  of  a  segment  of  a 

circular  spindle,  Fig.  62,  the  length  of  it  being  3.2495  inches, 
the  radius  of  its  arc  10,  the  central  distance  7-071,  and  the 
length  of  its  side,  id,  3.927  inches? 

3.2495  x  l0=S2A95  =  length  X  radius. 
3.927  X  7.071 =27.7678= length  of  arc  x  central  distance. 
32.495  —  27.7678 =4.7272 = difference  of  products. 
4.7272x2  =  9.4544x3.1416=29.702,  which  is  the  result  re- 
quired. 

Ex.  2.  The  segment  of  a  circular  spindle  is  14.142  feet  in 
length,  the  radius  of  its  arc  is  20,  and  the  distance  between 
the  plane  of  the  segment,  i  e,  and  the  centre  of  the  revolving 
arc,  o  e,  Fig.  62,  is  14.142,  and  the  length  of  its  side,  t  d,  is 
15.708  ;  what  is  its  convex  surface  1      Ans.  381.3745  feet. 


MENSURATION   OF   AREAS,  LINES,  AND   SURFACES.        123 

For  Surface  of  a  Circular  Spindle,  Zone,  or  Segment 

General  Formula.  S  =  2(lr—ac)p,  I  representing  length  of 
spindle,  segment,  or  zone,  a  the  length  of  its  revolving  arc,  r  the 
radius  of  the  generating  circle,  and  c  the  central  distance. 

Illustration. — The  length  of  a  circular  spindle  is  14.142 
inches,  th5  length  of  its  revolving  arc  is  15.708,  the  radius 
of  its  generating  circle  is  10,  and  the  distance  of  its  centre 
from  the  centre  of  the  circle  from  which  it  is  generated  is 
7.071 ;  what  is  its  surface? 

2  x  (14.142  x  10-15.708  x  7.071)  X  3.1416  =  190.6869  =  re- 
sult required. 

Centre  of  Gravity.     See  Appendix,  page  283. 

Note. — The  surface  of  the  frustrum  of  a  spindle  is  obtained  by  the 
division  of  the  surface  of  a  zone. 

Cycloidal  Spindle. 
To  ascertain  the  Convex  Surface  of  a  Cycloidal  Spindle,  Fig.  63. 
Rule. — Multiply  the  area  of  the  generating  circle  by  64, 
and  divide  it  by  3  ;  the  quotient  will  give  the  surface  re- 
quired. 


_     «X64 

\)r,  — - — —surface. 


Fig.  63. 


Example. — The  area  of  the  generating  circle,  a  be,  of  a 
cycloidal  spindle,  d  e,  is  32  inches ;  what  is  the  surface  of  the 
spindle  ? 


124        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

32  x  64 = 2048 =area  of  circle  x  64. 
2048 -r-3  =  682.667 —above  product-^- 3 —surface  required. 
Ex.  2.  The  diameter  of  the  generating  circle  of  a  cycloidal 
spindle  is  20.375  inches ;  what  is  the  surface  of  the  spindle? 

Ans.-  6955.7483  in. 
Ex.  3.  The  diameter  of  the  generating  circle  of  a  cycloidal 
spindle  is  14.5  inches ;  what  is  the  surface  of  the  spindle? 

Ans.  3522.773  in. 
Ex.  4.  The  radius  of  the  generating  circle  of  a  cycloidal 
spindle  is  8.5  inches;  what  is  the  surface  of  the  spindle  in 
square  feet !  Ans.  33.633  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

Note. — The  area  of  a  cycloidal  spindle  is  twice  the  area  of  the  cy- 
cloid, to  ascertain  which,  see  rule,  page  108. 


Elliptic,  Parabolic,  and  Hyperbolic  Spindles. 

The  rules  to  ascertain  the  surfaces  of  either  an  Elliptic,  Parabolic,  or 
Hyperbolic  Spindle,  or  of  zones  or  segments  of  them,  are  of  a  character 
to  preclude  their  being  given  in  such  a  form  as  would  be  consistent  with 
the  design  of  this  work ;  hence  they  are  omitted  here. 

See  Appendix,  p.  278,  279. 


Ellipsoid,  Paraboloid,  or  Hyperboloid  of  Revolution. 

Definition.  Figures  alike  to  a  cone  generated  by  the  revolu- 
tion of  a  conic  section  around  its  axis. 

Note. — These  figures  are  usually  known  as  Conoids. 

When  they  are  generated  by  the  revolution  of  an  ellipse, 
they  are  called  ellipsoids,  and  when  by  a  parabola,  parabo- 
loids, &c,  &c,  &c. 

The  revolution  of  an  arc  of  a  conic  section  around  the  axis  of  the 
curve  will  give  a  segment  of  a  conoid. 


MENSURATION   OF   AREAS,  LINES,  AND    SURFACES.        125 

Ellipsoid.* 

To  ascertain  the  Convex  Surface  of  an  Ellipsoid,  Fig.  64. 

Kule. — Add  together  the  square  of  the  base  a  b  and  four 
times  the  square  of  the  height  c  d ;  multiply  the  square  root 
of  half  their  sum  by  3.1416,  and  this  product  by  the  radius 
of  the  base.     The  product  will  give  the  surface  required. 

Or,  \J —  x  3.1416  X  r=zsurface,h  representing  the  height 

of  the  ellipsoid. 

Fig.  64.    s*r±. 


Example. — The  base  a  b  of  the  ellipsoid,  Fig.  64,  is  10 
inches,  and  the  height  c  d  7  ;  what  is  its  surface  ? 

102-f72  x  4  =  296—  sum  of  the  square  of  the  base  and  4  times 
the  square  of  the  height. 

296-f-2  =  148,  and  ^148  =  12.1655 —square  root  of  half 
the  above  sum. 

12.1655  x  3.1416  x-^r  =  191. 0957  =product  of  root   above 

obtained  x  3.1416,  and  that  product  by 'the  radius  of  the  base  = 
the  surface  required. 

Ex.  2.  The  base  a  b  of  an  ellipsoid  is  14  inches,  and  the 
height  c  d  5  ;  what  is  the  convex  surface  of  it  ? 

Ans.  267.534  in. 

To  ascertain  the  Convex  Surface  of  a  Segment,  Frusirum,  or  Zone 
of  an  Ellipsoid. 

See  rules  for  the  convex  surface  of  a  segment,  frustrum,  or 
zone  of  an  ellipsoid,  p.  93-95. 

*  An  ellipsoid  is  a  semi-spheroid.     (See  p.  91-94.) 


126        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

Paraboloid. 

To  ascertain  the  Convex  Surface  of  a  Paraboloid,  Fig.  65. 

Rule. — From  the  cube  of  the  square  root  of  the  sum  of  four 
times  the  square  of  the  height,  b  d,  and  the  square,  of  the  ra- 
dius of  the  base,  d  a,  subtract  the  cube  of  the  radius  of  the 
base ;  multiply  the  remainder  by  the  quotient  of  3.1416  times 
the  radius  of  the  base  divided  by  six  times  the  square  of  the 
height,  and  the  product  will  give  the  surface  required. 

rxp 


Or,  [(-\/4A2-f  r2)3— r3]  x  ~ — ^= convex  surface. 


d 

Example. — The  axis  b  d  of  a  paraboloid,  Fig.  65,  is  40 
inches,  the  radius  a  d  of  its  base  is  18  inches;  what  is  its 
convex  surface  ? 

402x4z=6400r=4  times  the  square  of  the  height. 

6400-t-182  =  6724=:sm«  of  the  above  product  and  the  square 
of  the  radius  of  the  base. 

(V6724)3-183=:545536=^  remainder' of  the  cube  of  the 
radius  of  the  base  subtracted  from  the  cube%  of  the  square  root  of 
the  preceding  sum. 

3.1416  x  18-^(6  x402)=.0058905=r^e  quotient  of  3.1416 
times  the  radius  of  the  base +-6  times  the  square  of  the  height. 

Hence  545536  x.0058905z=3213.48=Me  product  of  the 
above  remainder  and  the  preceding  quotient =the  result  required. 

Ex.  2.  The  axis  b  d  of  a  paraboloid  is  20  inches,  and  the 
diameter  of  its  base  a  c  is  60  inches ;  what  is  its  convex  sur- 
face? .  Ans.  3848.46  in. 


MENSURATION   OF   AREAS,  LINES,  AND    SURFACES.        127 


Ex.  3.  The  axis  of  a  parabolic  conoid  is  18  inches,  the  ra- 
dius of  its  base  40  inches ;  what  is  its  convex  surface  <? 

Ans.  5937.16  in. 
Centre  of  Gravity.-    See  Appendix,  p.  284. 

Any  Figure  of  Revolution. 

To  ascertain  the  Convex  Surface  of  any  Figure  of  Revolution, 
Figs.  66,  67,  and  68. 

Rule. — Multiply  the  length  of  the  generating  line  by  the 
circumference  described  by  its  centre  of  gravity,  and  the  prod- 
uct will  give  the  surface  required. 

Or,  lx%rxp— surface ;  r  representing  radius  of  centre  of 
gravity. 

Fig.  66.  e 


,  a 


-«f 


Example. — If  the  generating  line  a  c  of  the  cylinder  a  c  df 
10  inches  in  diameter,  Fig.  66,  is  10,  then  the  centre  of  grav- 
ity of  it  will  be  in  b,  the  radius  of  which  is  b  r—5. 

Hence  10x5  x 2  x3. 1416  =  314.16  =  ^e  convex  surface  of 
the  cylinder. 

Again,  If  the  generating  line  is  c  a  e  g,  and  it  is  (e  a=z5, 
a  c=10,  and  c  <7=5)=20,  then  the  centre  of  gravity  o  will 
be  in  the  middle  of  the  line  joining  the  centres  of  gravity  of 
the  triangles  e  a  c  and  a  c  #=3.75. 

Hence  20  x  3775x2  X  3.141 6  =  471.24  =zthe  entire  surface 
of  the  cylinder. 

C  Convex  surface  as  above 314.16 

Proof  ■<  Area  of  each  end,  102x-7854  =  78.54, 

(     and  78.54x2= 157.08 

471.24 


128        MENSURATION    OF    AREAS,  LINES,  AND    SURFACES. 
Fig.  67.  a 


Ex.  2.  If  the  generating  elements  of  a  cone,  Fig.  67,  are 
a  dz=\0,d  c=10,  and  a  c  the  generating  line  =  14.142,  the 
centre  of  gravity  of  which  is  in  o,  and  o  r=z5. 

Then,  14.142  x5  x2  x  3.1416=444.285=^6  convex  surf  ace 

2 

of  the  cone,  and  10x2  x. 7854=314. \Q>  —  area  of  base. 

Hence,  444.285-j-314.16  =  758.445  =  w/zo/6  surface  of  cone. 

Again.  If  the  generating  line  is  a  c  d— 24.142,  then  the 
centre  of  gravity  will  be  in  n,  in  the  middle  of  the  line,  join- 
ing the  angle  of  the  generating  line  and  the  base  a  d  at  r=5. 

Hence,  24.142  x 5~x2  X 3.1416 =758.445  —whole  surf  ace  of 
cone. 


Fig.  68 


Ex.  3.  If  the  generating  elements  of  a  sphere,  Fig.  68,  are 
a  c=10,  a  b  c  will  be  15.708,  the  centre  of  gravity  of  which 
is  in  o,  and  by  rule,  page  80,  o  r=3.183. 

Hence  15.708x3.183  x2x 3.1416 =314.16 -the surface  of 
tlie  sphere. 

To  ascertain  the  Area  of  an  Irregular  Figure. 

Rule. — Take  a  uniform  piece  of  board  or  pasteboard, 
weigh  it,  cut  out  the  figure  of  which  the  area  is  required,  and 
weigh  it ;  then,  as  the  weight  of  the  board  or  pasteboard  is  to 
the  entire  surface,  so  is  the  weight  of  the  figure  to  its  surface. 


MENSURATION   OP   AREAS,  LINES,  AND    SURFACES.        129 
CAPILLARY   TUBE. 

To  ascertain  the  Diameter  of  a  Capillary  Tube. 
Rule. — Weigh  the  tube  when  empty,  and  again  when  filled 
with  mercury ;  Subtract  the  one  weight  from  the  other ;  re- 
duce the  difference  to  troy  grains,  and  divide  it  by  the  length 
of  the  tube  in  inches.  Extract  the  square  root  of  this  quo- 
tient, and  multiply  it  by  .0192245,  and  the  product  will  be 
the  diameter  of  the  tube  in  inches. 

/  w 
Or,*/  —  x. 0192245  —diameter;   w   representing  difference 

in  weights  in  Troy  grains,  and  I  the  length  of  the  tube. 

Example. — The  difference  in  the  weights  of  a  capillary 
tube  when  empty  and  when  filled  with  mercury  is  90  grains, 
and  the  length  of  the  tube  is  10  inches ;  what  is  the  diameter 
of  it? 

90-t-lO  =  9 =weight  of mercury -+■  length  of  tube;  ^9  —  3,  and 
3  x  .0192245 =.0576735  =  the  square  root  of  the  above  quotient 
X  .0192245  =  diameter  of  tube  required. 

Proof. — The  weight  of  a  cubic  inch  of  mercury  is  3442.75 
Troy  grains,  and  the  diameter  of  a  circular  inch  of  equal  area 
to  a  square  inch  is  1.128  (p.  81). 

If,  then,  3442.75  grams  occupy  1  cubic  inch,  90  grains  will 
require  .0261419  cubic  inch,  which,  -^  10  for  the  height  of  the 
tube,  gives  .00261419  inch  for  the  area  of  the  section  of  the  tube. 

Then  v'.00261419  =  .051129=SKfeo/Me  square  of  a  column 
of  mercury  of  this  area. 

Hence  .051129  x  1.128,  which  is  the  ratio  between  the  side  of 
a  square  and  the  diameter  of  a  circle  of  equal  area  —  .057 '67 '35. 

F2 


130        MENSURATION   OF   AREAS,  LINES,  AND   SURFACES. 


LENGTHS    OF    CIRCULAR   ARCS. 


Table  of  the  Lengths  of  Circular  Arcs,  the  Diameter  of  a  Circle 
being  Unity,  and  assumed  to  be  divided  into  1000  equal  Parts. 


Height. 

Length. 

Height. 

Length. 

Height. 

Length. 

Height. 

Length. 

.100 

1.0265 

.134 

1.0472 

.168 

1.0737 

.202 

1.1055 

.101 

1.0270 

.135 

1.0479 

.169 

1.0745 

.203 

1.1065 

.102 

1.0275 

.136 

1.0486 

.170 

1.0754 

.204 

1.1075 

.103 

1.0281 

.137 

1.0493 

.171 

1.0762 

:'205 

1.1085 

.104 

1.0286 

.138 

1.0500 

.172    1.0771    .206 

1.109G 

.105 

1.0291 

.139 

1.0508 

.173 

1.0780 

.207 

1.1106 

.106 

1.0297 

.140 

1.0515 

.174 

1/0789 

.208 

1.1117 

.107 

1.0303 

.141 

1.0522 

.175 

1.0798 

.209 

1.1127 

.108 

1.0308 

.142 

1.0529 

-.176 

1.0807 

.210 

1.1137 

.109 

1.0314 

.143 

1.0537 

.177 

1.0816 

.211    1.1148 

.110 

1.0320 

.144 

1.0544 

.178 

1.0825 

.212 

1.1158 

.111 

1.0325- 

.145 

1.0552 

.179 

1.0834 

.213 

1.1169 

.112 

1.0331 

.146 

1.0559 

.180 

1.0843 

.214   1.1180 

.113 

1.0337 

.147 

1.0567 

.181 

1.0852    .215;1.1190 

.114 

1.0343 

.148 

1.0574 

.182 

1.0861 

.21611.1201 

.115 

1.0349 

.149 

1.0582 

.183 

1.0870 

.217  1 1.1212 

.116  1 1.0355 

.150 

1.0590 

.184 

1.0880 

.218 

1.1223 

.11711.0361 

.151 

1.0597 

.185 

1.0889 

.219 

1.1233 

.118    1.0367 

.152 

1.0605 

.186 

1.0898 

.220   1.1245 

.119    1.0373 

.153 

1.0613 

.187 

1.0908 

.221 |1.1256 

.120 ; 1.0380 

.154 

1.0621 

.188 

1.0917 

.222 

1.1266 

.12111.0386 

.155 

1.0629 

.189    1.0927 

.223 

1.1277 

.122    1.0392 

.156 

1.0637 

.190   1.0936 

.224 

1.1289 

.123 

1.0399 

.157 

1.0645 

.191 

1.0946 

.225 

1.1300 

.124 

1.0405 

.158 

1.0653 

.192 

1.0956 

.226 

1.1311 

.125 

1.0412 

.159 

1.0661 

.193 

1.0965 

.227 

1.1322 

.126 

1.0418 

.160 

1.0669 

.194 

1.0975 

.228  11.1333 

.127 

1.0425 

.161 

1.0678 

.195 

1.0985 

.229    1.1344 

.128 

1.0431 

.162 

1.0686 

.196 

1.0995 

.230; 1.1356 

.129 

1.0438 

.163 

1.0694 

.197 

1.10051.231 ! 1.1367 

.130 

1.0445 

.164 
.165 

1*0703 

.198 

1.1015S. 232    1.1379 

.131 

1.0452 

1.0711 

.199 

1.10251.233    1.1390 

.132 

1.0458 

.166 

1.0719 

.200 

1.1035    .234    1.1402 

.133 

1.0465 

.167 

1.0728 

.201 

1.10451 

.235 

1.1414 

MENSURATION    OF  AREAS,  LINES,  AND  SURFACES.         131 


Table  of  Circular  Arcs — Continued. 


Height. 

Length.    1 

Height. 

Length. 

Height. 

Length. 

Height 

Length. 

.236 

1.1425 

.274 

1.1897 

.312 

1.2422 

.350 

1.3000 

.237 

1.1436 

.275 

1.1908 

.313 

1.2436 

.351 

1.3016 

.238 

1.1448 

.276 

1.1921 

.314 

1.2451 

.352 

1.3032 

.239 

1.1460 

.277 

1.1934 

.315 

1.2465 

.353 

1.3047 

.240 

1.1471 

.278 

1.1948 

.316 

1.2480 

.354 

1.3063 

.241 

1.1483 

.279 

1.1961 

.317 

1.2495 

.355 

1.3079 

.242 

1.1495 

.280 

1.1974 

.318 

1.2510 

.356 

1.3095 

.243 

1.1507 

.281 

1.1989 

.319 

1.2524 

.357 

1.3112 

.244 

1.1519 

.282 

1.2001 

.320 

1.2539 

.358 

1.3128 

.245 

1.1531 

.283 

1.2015 

.321 

1.2554 

.359 

1.3144 

.246 

1.1543 

.284 

1.2028 

.322 

1.2569 

.360 

1.3160 

.247 

1.1555 

.285 

1.2042 

.323 

1.2584 

.361 

1.3176 

.248 

1.1567 

.286 

1.2056 

.324 

1.2599 

.362 

1.3192 

.249 

1.1579 

.287 

1.2070 

.325 

1.2614 

.363 

1.3209 

.250 

1.1591 

.288 

1.2083 

.326 

1.2629 

.364 

1.3225 

.251 

1.1603 

.289 

1.2097 

.327 

1.2644 

.365 

1.3241 

.252 

1.1616 

.290 

1.2120 

.328 

1.2659 

.366 

1.3258 

.253 

1.1628 

.291 

1.2124 

.329    1.2674 

.367 

1.3274 

.254 

1.1640 

.292 

1:2138 

.330 

1.2689 

.368 

1.3291 

.255 

1.1653 

.293 

1.2152 

.331 

1.2704 

.369 

1.3307 

.256 

1.1665 

.294 

1.2166 

.332 

1.2720 

.370    1.3323 

.257 

1.1677 

.295 

1.2179 

.333 

1.2735 

.371 i 1.3340 

.258 

1.1690 

.296 

1.2193 

.334 

1.2750 

.372 ! 1.3356 

.259    1.1702 

.297 

1.2206 

.335 

1.2766 

.373    1.3373 

.260    1.1715 

.298 

1.2220 

.336 

1.2781 

.374    1.3390 

.261 

1.1728 

.299 

1.2235 

.337 
.338 

1.2796 

.375    1.3406 

.262 

1.1740 

.300 

1.2250 

1.2812 

.376  11.3423 

.263 

1.1753 

.301 

1.2264 

.339 

1.2827 

.377 i 1.3440 

.264 

1.1766 

.302 

1.2278 

.340 

1.2843 

.378 

1.3456 

.265 

1.1778 

.303 

1.2292 

.341 

1.2858 

.379 

1.3473 

.266 

1.1791 

.304 

1.2306 

.342 

1.2874 

.380 

1.3490 

.267 

1.1804 

.305 

1.2321 

.343 

1.2890 

.381 

1.3507 

.268 

1.1816 

.306 

1.2335 

.344 

1.2905 

.382 

1.3524 

.269 

1.1829 

.307 

1.2349 

.345 

1.2921 

.383 

1.3541 

.270 

1.1843 

.308 

1.2364 

.346 

1.2937 

.384 

1.3558 

.271 

1.1856 

.309 

1.2378 

.347 

1.2952 

.385 

1.3574 

.272 

1.1869 

.310 

1.2393 

.348 

1.2968 

.386 

1.3591 

.273 

1.1882 

.311 

1.2407 

.349 

1.2984 

.387 

1.3608 

132        MENSURATION   OF  AREAS,  LINES,  AND   SURFACES. 


Table  of 

Circular  Arcs — 

■Continued. 

Height.  |     Length. 

Height.  |     Length. 

Height.  |     Length.      Height.  |    Length. 

.388    1.3625 

.417 

1.4132 

.445 

1.4644 

.473 

1.5176 

.389 

1.3643 

.418 

1.4150 

.446 

1.4663 

.474 

1.5196 

.390 

1.3660 

.419 

1.4168 

.447 

1.4682 

.475 

1.5215 

.391 

1.3677 

.420 

1.4186 

.448 

1.4700 

.476 

1.5235 

.392 

1.3694 

.421 

1.4204 

.449 

1.4719 

.477 

1.5254 

.393    1.3711 

.422 

1.4222 

.450 

1.4738 

.478 

1.5274 

.394; 1.3728 

.423 

1.4240 

.451 

1.4757 

.479 

1.5293 

.395    1.3746 

.424 

1.4258 

.452 

1.4775 

.480 

1.5313 

.396 

1.3763 

.425 

1.4276 

.453 

1.4794 

.481 

1.5332 

.397 

1.3780 

.426 

1.4295 

.454 

1.4813 

.482 

1.5352 

.398 

1.3797 

.427 

1.4313 

.455 

1.4832 

.483 

1.5371 

.399 

1.3815 

.428 

1.4331 

.456 

1.4851 

.484 

1.5391 

.400 

1.3832 

.429 

1.4349 

.457 

1.4870 

.485 

1.5411 

.401 

1.3850 

.430 

1.4367 

.458 

1.4889 

.486 

1.5430 

.402 

1.3867 

.431 

1.4386 

.459 

1.4908 

.487 

1.5450 

.403 

1.3885 

.432 

1.4404 

.460 

1.4927 

.488 

1.5470 

.404 

1.3902 

.433 

1.4422 

.461 

1.4946 

.489 

1.5489 

.405 

1.3920 

.434 

1.4441 

.462 

1.4965 

.490 

1.5509 

.406 

1.3937 

.435 

1.4459 

.463 

1.4984 

.491 

•1.5529 

.407 

1.3955 

.436 

1.4477 

.464 

1.5003 

.492 

1.5549 

.408 

1.3972 

.437 

1.4496 

.465 

1.5022 

.493 

1.5569 

.409 

1.3990 

.438 

1.4514 

.466 

1.5042 

.494 

1.5585 

.410 

1.4008 

.439 

1.4533 

.467 

1.5061 

.495 

1.5608 

.411 

1.4025 

.440 

1.4551 

.468 

1.5080 

.496 

1.5628 

.412 

1.4043 

.441 

1.4570 

.469 

1.5099 

.497 

1.5648 

.413 

1.4061 

.442 

1.4588 

.470 

1.5119 

.498 

1.5668 

.414 

1.4079 

.443 

1.4607 

.471 

1.5138 

.499 

1.5688 

.415 

1.4097 

.444 

1.4626 

.472 

1.5157 

.500 

1.5708 

.416 

1.4115 

To  find  the  Length  of  an  Arc  of  a  Circle  by  the  foregoing  Table. 

Rule. — Divide  the  height  by  the  base,  find  the  quotient  in 
the  column  of  heights,  and  take  the  length  of  that  height  from 
the  next  right-hand  column.  Multiply  the  length  thus  ob- 
tained by  the  base  of  the  arc,  and  the  product  will  be  the 
length  of  the  arc  required. 


MENSURATION    OF   AREAS,  LINES,  AND    SURFACES.        133 

Example. — What  is  the  length  of  an  arc  of  a  circle,  the 
span  or  base  being  100  feet,  and  the  height  25  feet? 

25  -f- 100  =  .25,  and  .25,  per  table,  =1.1591,  which,  multiplied 
by  100,  =1 15,9100 /ee*. 

Note. —  When,  in  the  division  of  a  height  by  the  base,  the  quo- 
tient has  a  remainder  after  the  third  place  of  decimals,  and  great 
accuracy  is  required, 

Take  the  length  for  the  first  three  figures,  subtract  it  from 
the  next  following  length,  multiply  the  remainder  by  the  said 
fraction,  and  add  the  product  to  the  first  length ;  the  sum  will 
be  the  length  for  the  whole  quotient. 

Example. — What  is  the  length  of  an  arc  of  a  circle,  the 
base  of  which  is  35  feet,  and  the  height  or  versed  sine  8  feet  ? 

8^35=.2285714;  the  tabular  length  for  .228  =  1. 1333,  and 
for  .229  =  1.1344,  the  difference  between  which  is  .0011.  Then, 
.5714  x  .0011 =.00062854.  i  / 

Hence,  .228  =  1.1333 

.0005714=  .00062854 

1.13392854,  the  sum  by  which  the  base 
of  the  arc  is  to  be  multiplied;  and  1.13392854  x  35  =  39  feet 
.6874989,  which,  x  12  for  inches =8. 25,  making  the  length  of  the 
arc  39  feet  8.25  inches. 


134        MENSURATION    OF  AREAS,  LINES,  AND    SURFACES. 


AREAS    OF    SEGMENTS    OF    A    CIRCLE. 

Table  of  the  Areas  of  the  Segments  of  a  Circle,  the  Diameter  of 
which  is  Unity,  and  assumed  to  be  divided  into  1000  equal 
Parts. 


Versed 
Bine. 

Seg.    Area. 

Versed 
sine. 

Seg.    Area. 

Versed 
sine. 

Seg.    Area. 

Versed 
sine. 

Seg.    Area. 

.001 

.00004 

.034 

.00827 

.067 

.02265 

.100 

.04087 

..002 

.00012 

.035 

.00864 

.068 

.02315 

.101 

.04148 

.003 

.00022 

.036 

.00901 

.069 

.02336 

.102 

.04208 

.004 

.00034 

.037 

.00938 

.070 

.02417 

.1(53 

.04269 

.005 

.00047 

.038 

.00976 

.071 

.02468 

.104 

.04310 

.006 

.00062 

.039 

.01015 

.072 

.02519 

.105 

.04391 

.007 

.00078 

.040 

.01054 

.073 

.02571 

.106 

.04452 

.008 

.00095 

.041 

.01093 

.074 

.02624 

.107 

.04514 

.009 

.00113 

.042 

.01133 

.075 

.02676 

.108 

.04575 

.010 

.00133 

.043 

.01173 

.076 

.02729 

.109 

.04638 

.011 

.00153 

.044 

.01214 

.077 

.02782 

.110 

.04700 

.012 

.00175 

.045 

.01255 

.078 

.02835 

.111 

.04763 

.013 

.00197 

.046 

.01297 

.079 

.02889 

.112 

.04826 

.014 

.00220 

.047 

.01339 

.080 

.02943 

.113 

.04889 

.015 

.00244 

.048 

.01382 

.081 

.02997 

.114 

.04953 

.016 

.90268 

.049 

.01425 

.082 

.03052 

.115 

.05016 

.017 

.00294 

.050 

.01468 

.083 

.03107 

.116 

.05080 

.018 

.00320 

.051 

.01512 

.084 

.03162 

.117 

.05145 

.019 

.00347 

.052 

.01556 

.085 

.03218 

.118 

.05209 

.020 

.00375 

.053 

.01601 

.086 

..03274 

.119 

.05274 

.021 

.00403 

.054 

.01646 

.087 

.03330 

.120 

.05338 

.022 

.00432 

.055 

•01691 

.088 
.089 

.03387 

.121 

.05404 

.023 

.00462 

.056 

.01737 

.03444 

.122 

.05469 

.024 

.00492 

.057 

.01783 

.090 

.03501 

.123 

.05534 

.025 

.00523 

.058 

.01830 

.091 

.03558 

.124 

.05600 

.026 

.00555 

.059 

.01877 

.092 

.03616 

.125 

.05666 

.027 

.00587 

.060 

.01924 

.093 

.03674 

.126 

.05733 

.028 

.00619 

.061 

.01972 

.094 

.03732 

.127 

.05799 

.029 

.00653 

.062 

.02020 

.095 

.03790 

.128 

.05866 

.030 

.00686 

.063 

•02068 

.096 

.03849 

.129 

.05933 

.031 

.00721 

.064 

.02117 

-097 

.03908 

.130 

.06000 

.032 

.00756 

.065 

.02165 

.098 

.03968 

.131 

.06067. 

.033 

.00791 

.066 

.02215 

.099 

•04027 

.132 

.06135 

MENSURATION    OF   AREAS,  LINES,  AND    SURFACES.         135 


Table  of  Areas  of  Segments  of  a  Circle — Continued. 


Versed 
sine. 

Seg.    Area. 

Versed 
sine. 

Seg.    Area, 

1  Versed 
1  sine. 

Seg.    Area. 

Versed 
sine. 

Seg.    Area. 

.133 

.06203 

.171 

.08929 

.209 

.11908 

.247 

.15095 

.134 

.06271 

.172 

.09004 

.210 

.11990 

.248 

.15182 

.135- 

.06339 

.173 

.09080 

.211 

.12071 

.249 

.15268 

.136 

.06407 

.174 

.09155 

.212 

.12153 

.250 

.15355 

.137 

.06476 

.175 

.09231 

.213 

.12235 

.251 

.15441 

.138 

.06545 

.176 

.09307 

.214 

.12317 

.252 

.15528 

.139 

.06614 

.177 

.09384 

.215 

.12399 

1.253 

.15615 

.140 

.06683 

.178 

.09460 

.216 

.12481 

.254 

.15702 

.141 

.06753 

.179 

.09537 

.217 

.12563 

.255 

.15789 

.142 

.06822 

.180 

.09613 

.218 

.12646 

.256 

.15876 

.143 

.06892 

.181 

.09690 

.219 

.12728 

.257 

.15964 

.144 

.06962 

.182 

.09767 

.220 

.12811 

.258 

.16051 

.145 

.07033 

.183 

.09845 

.221 

.12894 

.259 

.16139 

.146 

.07103 

.184 

.09922 

.222 

.12977 

.260 

.16226  * 

.147 

.07174 

.185 

. 10000 

.223 

.13060 

.261 

.16314 

.148 

.07245 

.186 

.10077 

.224 

.13144 

.262 

. 16402 

.149 

.07316 

.187 

.10155 

.225 

.13227 

.263 

.16490 

.150 

.07387 

.188 

.10233 

.226 

.13311 

.264 

.16578 

.151 

.07459 

.189 

.10312 

.227 

.13394 

.265 

.16666 

.152 

.07531 

.190 

.10390 

.228 

.13478 

.266 

.16755 

.153 

.07603 

.191 

.10468 

.229 

.13562 

.267 

.16844 

.154 

.07675 

.192 

. 10547 

.230 

.13646 

.268 

.16931 

.155 

.07747 

.193 

.10626 

.231 

.13731 

.269 

.17020 

.156 

.07820 

.194 

.10705 

.232 

.13815 

.270 

.17109 

.157 

.07892 

.195 

.10784 

.233 

.13900 

.271 

.17197 

.158 

.07965 

.196 

.10864 

.234 

.13984 

.272 

.17287 

.159 

.08038 

.197 

.10943 

.235 

.14069 

.273 

.17376 

.160 

.08111 

.198 

.11023 

.236 

.14154 

.274 

.17465 

.161 

.08185 

.199 

.11102 

.237 

.14239 

.275 

.17554 

.162 

.08258 

.200 

.11182 

.238 

.14324 

.276 

.17643 

.163 

.08332 

.201 

.11262 

.239 

.14409 

.277 

.17733  4 

.164 

.08406 

.202 

.11343 

.240 

.14494 

.278 

.17822 

.165 

.08480 

.203 

.11423 

.241 

.14580 

.279 

.17912 

.166 

.08554 

.204 

.11503 

.242 

.14665 

.280 

.18002 

.167 

.08629 

.205 

.11584 

.243 

.14751 

.281 

.18092 

.168 

.08704 

.206 

.11665 

.244 

.14837 

.282 

.18182 

.169 

,08779 

.207 

.11746 

.245 

.14923 

.283 

.18272 

.170 

.08853 

.208 

.11827  .246  .15009 

THE  ^\ 

.284 

.18361 

I   U 

NIVFl 

3-21TV 

136        MENSURATION   OF   AREAS,  LINES,  AND    SURFACES. 


Table  of  Areas  of  Segments  of  a  Circle — Continued. 


Versed 
sine. 

Seg.    Area. 

Versed 
sine. 

Seg.    Area 

Versed 
sine. 

.285 

.18452 

.323 

.21947 

.361 

.286 

.18542 

.324 

.22040 

.36? 

.287 

.18633 

.325 

.22134 

.363 

.288 

.18723 

.326 

.22228' 

.364 

.289 

.18814 

.327 

.22321 

.365 

.290 

.18905 

.328 

.22415 

.366 

.291 

.18995 

.329 

.22509 

.367 

.292 

.19086 

.330 

.22603 

.368 

.293 

.19177 

.331 

.22697 

.369 

.294 

.19268 

.332 

.22791 

.370 

.295 

.19360 

.333 

.22886 

.371 

.296 

.19451 

.334 

.22980 

.372 

.297 

.19542 

.335 

.23074 

.373 

..298 

.19634 

.336 

.23169 

.374 

.299 

.19725 

.337 

.23263 

.375 

.300 

.19817 

.338 

.23358 

.376 

.301 

.19908 

.339 

.23453 

.377 

.302 

.20000 

.340 

.23547 

.378 

.303 

.20092 

.341 

.23642 

.379 

.304 

.20184 

.342 

.23737 

.380 

.305 

.20276 

.343 

.23832 

.381 

.306 

.20368 

.344 

.23927 

.382 

.307 

.20460 

.345 

.24022 

.383 

.308 

.20553 

.346 

.24117 

.384 

.309 

.20645 

.347 

.24212 

.385 

.310 

.20738 

.348 

.24307 

.386 

.311 

.20830 

.349 

.24403 

.387 

.312 

.20923 

.350 

.24498 

.388 

.313 

.21015 

.351 

.24593 

.389 

.314 

.21108 

.352 

.24689 

.390 

.315 

.21201 

.353 

.24784 

.391 

.316 

.21294 

.354 

.24880 

.392 

.317 

.21387 

.355 

.24976 

.393 

.318 

.21480 

.356 

.25071 

.394 

.319 

.21573 

.357 

.25167 

.395 

.320 

.21667 

.358 

.25263 

.396 

.321 

.21760 

.359 

.25359 

.397 

.322 

.21853 

.360 

.25455 

.398 

Seg.    Area 

Versed 
sine. 

Seg.    Area. 

.25551 

.399 

.29239 

.25647 

.400 

.29337 

.25743 

.401 

.29435 

.25839 

.402 

.29533 

.25936 

.403 

.29631 

.26032 

.404 

.29729 

.26128 

.405 

.29827 

.26225 

.406 

.29925 

.26321 

.407 

.30024 

.26418 

.408 

.30122 

.26514 

.409 

.30220 

.26611 

.410 

.30319 

.26708 

.411 

.30417 

.26804 

.412 

.30515 

.26901 

.413 

.30614 

.26998 

.414 

.30712 

.27095 

.415 

.30811 

.27192 

.416 

.30909 

.27289 

.417 

.31008 

.27386 

.418 

.31107 

.27483 

.419 

.31205 

.27580 

.420 

.31304 

.27677 

.421 

.31403 

.27775 

.422 

.31502 

.27872 

.423 

.31600 

.27969 

.424 

.31699 

.28067 

.425 

.31798 

.28164 

.426 

.31897 

.28262 

.427 

.31996 

.28359 

.428 

.32095 

.28457 

.429 

.32194 

.28554 

.430 

.32293 

.28652 

.431 

.32391 

.28750 

.432 

.32490 

.28848 

.433 

.32590 

.28945 

.434 

.32689 

.29043 

.435 

.32788 

.29141 

.436 

.32887 

MENSURATION   OF  AREAS,  LINES,  AND   SURFACES.         137 


Table,  of  Areas  c 

)f  Segments  of  a 

Circle — ( 

Continued. 

Versed                             Versed                             Versed 

-,           ,            Versed 

sine. 

Seg.    Area. 

sine. 

Seg.    Area. 

sine. 

Seg.    Area. 

sine. 

Seg.    Area. 

.437 

.32987 

.453 

.34577 

.469 

.36172 

.485 

.37770 

.438 

.33086 

.454 

.34676 

.470 

.36272 

.486 

.37870 

.439 

.33185 

.455 

.34776 

.471 

.36371 

.487 

.37970 

.440 

.33284 

.456 

.34875 

.472 

.36471 

.488 

.38070 

.441 

.33384 

.457 

.34975 

.473 

.36571 

.489 

.38170 

.442 

.33483 

.458 

.35075 

.474 

.36671 

.490 

.38270 

.443 

.33582 

.459 

.35174 

.475 

.36781 

.491 

.38370 

.444 

.33682 

.460 

.35274 

.476 

.36871 

.492 

.38470 

.445 

.33781 

.461 

.35374 

.477 

.36971 

.493 

.38570 

.446 

.33880 

.462 

.35474 

.478 

.37071 

.494 

.38670 

.447 

.33980 

.463 

.35573 

.479 

.37170 

.495 

.38770 

.448 

.34079 

.464 

.35673 

.480 

.37276 

.496 

.38870 

.449 

.34179 

.465 

.35773 

.481 

.37370 

.497 

.38970 

.450 

.34278 

.466 

.35872 

.482 

.37470 

.498 

.39070 

.451 

.34378 

.467 

.35972 

.483 

.37570 

.499 

.39170 

.452 

.34477 

.468 

.36072 

.484 

.37670 

.500 

.39270 

To  find  the  Area  of  a  Segment  of  a  Circle  by  the  above  Table. 

Rule. — Divide  the  height  or  versed  sine  by  the  diameter  of 
the  circle,  and  find  the  quotient  in  the  column  of  versed  sines. 
Take  the  area  noted  in  the  next  column,  and  multiply  it  by 
the  square  of  the  diameter,  and  it  will  give  the  area  required. 

Example. — Required  the  area  of  a  segment,  its  height  be- 
ing lO,  and  the  diameter  of  the  circle  50  feet. 

10  +  50  =  .2,  and  .2,  per  table,  =.11182;  then,  .11182  x502 
=279.55feet. 

Note. —  When,  in  the  division  of  a  height  by  the  base,  the  quo- 
tient has  a  remainder  after  the  third  place  of  decimals,  and  great 
accuracy  is  required, 

Take  the  length  for  the  first  three  figures,  subtract  it  from 
the  next  following  length,  multiply  the  remainder  by  the  said 
fraction,  and  add  the  product  to  the  first  length ;  the  sum  will 
be  the  length  for  the  whole  quotient. 


138        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

Example. — What  is  the  area  of  a,  segment  of  a  circle,  the 
diameter  of  which  is  10  feet,  and  the  height  of  it  1.575  feet? 

1.575-f- 10=.  1575  ;  the  tabular  area  for  .157=.07892,  and 
for  .15.8  =  .07964,  the  difference  between  which  is  .00072. 

Then,  .5  x  .00072  =  .000360. 

Hence,  .157   =.07892 

.0005  =  .00036 

.07928,  the  sum  by  which  the  square 
of  the  diameter  of  the  circle  is  to  be  multiplied;  and  .07928  x  102 
= 7.928 /£<*. 


MENSURATION    OF   AREAS,  LINES,  AND    SURFACES.        139 


AREAS   OF   THE   ZONES    OF   A    CIRCLE. 

Table  of  the  Areas  of  the  Zones  of  a  Circle,  the  Diameter  of 
which  is  Unity,  and  assumed  to  be  divided  into  1000  equal 
Parts. 


Height. 

Area. 

Height. 

Area. 

Height. 

Area, 

Height. 

Area. 

.001 

.00100 

.034 

.03397 

.067 

.06680 

.100 

.09933 

.002 

.00200 

.035 

.03497 

.068 

.06780 

.101 

.10031 

.003 

.00300 

.036 

.03597 

.069 

.06878 

.102 

.10129 

.004 

.00400 

.037 

.03697 

.070 

.06977 

.103 

.10227 

.005 

.00500 

.038 

.03796 

.071 

.07076 

.104 

.10325 

.006 

.00600 

.039 

.03896 

.072 

.07175 

.105 

.10422 

.007 

.00700 

.040 

.03996 

.073 

.07274 

.106 

.10520 

.008 

.00800 

.041 

.04095 

.074 

.07373 

.107 

.10618 

.009 

.00900 

.042 

.04195 

.075 

.07472 

.108 

.10715 

.010 

.01000 

.043 

.04295 

.076 

.07570 

.109 

.10813 

.011 

.01100 

.044 

.04394 

.077 

.07669 

.110 

.10911 

.012 

.01200 

.045 

.04494 

.078 

.07768 

.111 

.11008 

.013 

.01300 

.046 

.04593 

.079 

.07867 

.112 

.11106 

.014 

.01400 

.047 

.04693 

.080 

.07966 

.113 

.11203 

.015 

.01500 

.048 

.04793 

.081 

.08064 

.114 

.11300 

.016 

.01600 

.049 

.04892 

..082 

.08163 

.115 

.11398 

.017 

.01700 

.050 

.04992 

.083 

.08262 

.116 

.11495 

.018 

.01800 

.051 

.05091 

.084 

.08360 

.117 

.11592 

.019 

.01900 

.052 

.05190 

.085 

.08459 

.118 

.11690 

.020 

.02000 

.053 

.05290 

.086 

.08557 

.119 

.11787 

.021 

.02100 

.054 

.05389 

.087 

.08656 

.120 

.11884 

.022 

.02200 

.055 

.05489 

.088 

.08754 

.121 

.11981 

.023 

.02300 

.056 

.05588 

.089 

.08853 

.122 

.12078 

.024 

.02400 

.057 

.05688 

.090 

.08951 

.123 

.12175 

.025 

.02500 

.058 

.05787 

.091 

.09050 

.124 

.12272 

.026 

.02599 

.059 

.05886 

.092 

.09148 

.125 

.12369 

.027 

.02699 

.060 

.05986 

.093 

.09246 

.126 

.12469 

.028 

.02799 

.061 

.06085 

.094 

.09344 

.127 

.12562 

.029 

.02898 

.062 

.06184 

.095 

.09443 

.128 

.12659 

.030 

.02998 

.063 

.06283 

.096 

.09540 

.129 

.12755 

.031 

.03098 

.064 

.06382 

.097 

.09639 

.130 

.12852 

.032 

.03198 

.065 

.06482 

.098 

.09737 

.131 

.12949 

.033 

.03298 

.066 

.06580 

.099 

.09835 

.132 

. 13045 

140        MENSURATION   OP  AREAS,  LINES,  AND  SURFACES. 


Table  of  the  Zones  of  a  Circle — Continued. 


Height. 

Area. 

Height. 

Area. 

Height 

Area. 

Height. 

Area. 

.133 

.13141 

.171 

.16761 

.209 

.20274 

.247 

.2365 

.134 

.13238 

.172 

.16855 

.210 

.20365 

.248 

.2374 

.135 

.13334 

.173 

.16948 

.211 

.20456 

.249 

.2382 

.136 

.13430 

.174 

.17042 

.212 

.20546 

.250 

.2391 

.137 

.13527 

.175 

.17136 

.213 

.20637 

.251 

.2400 

.138 

.13623 

.176 

.17230 

.214 

.20727 

.252 

.2408 

.139 

.13719 

.177 

.17323 

.215 

.20818 

.253 

.2417 

.140 

.13815 

.178 

.17417 

.216 

.20908 

.254 

.2426 

.141 

.13911 

.179 

.17510 

.217 

.20998 

.255 

.2434 

.142' 

.14007 

.180 

.17603 

.218 

.21088 

.256 

.2443 

.143 

.14103 

.181 

.17697 

.219 

.21178 

.257 

.2451 

.144 

.14198 

.182 

.17790 

.220 

.21268 

.258 

.2460 

.145 

.14294 

.183 

.17883 

.221 

.21358 

.259 

.2469 

.146 

.14390 

.184 

.17976 

.222 

.21447 

.260 

.2477 

.147 

. 14485 

.185 

.18069 

.223 

.21537 

.261 

.2486 

.148 

.14581 

.186 

.18162 

.224 

.21626 

.262 

.2494 

.149 

.14677 

.187 

.18254 

.225 

.21716 

.263 

.2502 

.150 

.14772 

.188 

.18347 

.226 

.21805 

.264 

.2511 

.151 

.14867 

.189 

.18440 

.227 

.21894 

.265 

.2520 

.152 

.14962 

.190 

.18532 

.228 

.21983 

.266 

.2528 

.153 

.15058 

.191 

.18625 

.229 

.22072 

.267 

.2537 

.154 

.15153 

.192 

.18717 

.230 

.22161 

.268 

.2545 

.155 

.15248 

.193 

.18809 

.231 

.22250 

.269 

.2553 

.156 

.15343 

.194 

.18902 

.232 

.22335 

.270 

.2562 

.157 

.15438 

.195 

.18994 

.233 

.22427 

.271 

.2570 

.158 

.15533 

.196 

.19086 

.234 

.22515 

.272 

.2579 

.159 

.15628 

.197 

.19178 

.235 

.22604 

.273 

.2587 

.160 

.15723 

.198 

.19270 

.236 

.22692 

.274 

.2595 

.161 

.15817 

.199 

.19361 

.237 

.22780 

.275 

.2604 

.162 

.15912 

.200 

.19453 

.238 

.22868 

.276 

.2612 

.163 

.16006 

.201 

.19545 

.239 

.22956 

.277 

.2620 

.164 

.16101 

.202 

.19636 

.240 

.23044 

.278 

.2629 

.165 

.16195 

.203 

.19728 

.241 

.23131 

.279 

.2637 

.166 

.16290 

.204 

.19819 

.242 

.23219 

.280 

.2645 

.167 

.16384 

.205 

.19910 

.243 

.23306 

.281 

.2654 

.168 

.16478 

.206 

.20001 

.244 

.23394 

.282 

.2662 

.169 

.16572 

.207 

.20092 

.245 

.23481 

.283 

.2670 

.170 

.16667 

.208 

.20183 

.246 

.23568 

.284 

.2678 

MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 


141 


Table 

of  the  Zones  oj 

a  Cm 

lie — Continued. 

Height.]   Area.   |  Height. 

Area. 

Height. 

Area.   1 

Height. 

Area. 

.285 

.26871 

.323 

.29886 

.361 

.32656 

.399 

.35122 

.286 

.26953 

.324 

.29962 

.362 

.32725 

.400 

.35182 

.287 

.27035 

.325 

.30039 

.363 

.32794 

.401 

.35242 

.288 

.27117 

.326 

.30114 

.364 

.32862 

.402 

.35302 

.289 

.27199 

.327 

.30190 

.365 

.32931 

.403 

.35361 

.290 

.27280' 

.328 

.30266 

.366 

.32999 

.404 

.35420 

.291 

.27362 

.329 

.30341 

.367 

.33067 

.405 

.35479 

.292 

.27443 

.330 

.30416 

.368 

.33135 

.406 

.35538 

.293 

.27524 

.331 

.30491 

.369 

.33203 

.407 

.35596 

.294 

.27605 

.332 

.30566 

.370 

.33270 

.408 

.35654 

.295 

.27686 

.333 

.30641 

.371 

.33337 

.409 

.35711 

.296 

.27766 

.334 

.30715 

.372 

.33404 

.410 

.35769 

.297 

.27847 

.335 

.30790 

.373 

.33471 

.411 

.35826 

.298 

.27927 

.336 

.30864 

.374 

.33537 

.412 

.35883 

.299 

.28007 

.337 

.30938 

.375 

.33604 

.413 

.35939 

.300 

.28088 

.338 

.31012 

.376 

.33670 

.414 

.35995 

.301 

.28167 

.339 

.31085 

.377 

.33735 

.415 

.36051 

.302 

.28247 

.340 

.31159 

.378 

.33801 

.416 

.36107 

.303 

.28327 

.341 

.31232 

.379 

.33866 

.417 

.36162 

.304 

.28406 

.342 

.31305 

.380 

.33931 

.418 

.36217 

.305 

.28486 

.343 

.31378 

.381 

.33996 

.419 

.36272 

.300 

.28565 

.344 

.31450 

.382 

.34061 

.420 

.36326 

.307 

.28644 

.345 

.31523 

.383 

.3^125 

.421 

.36380 

.308 

.28723 

.346 

.§1595 

.384 

.34190 

.422 

.36434 

.309 

.28801 

.347 

.31667 

.385 

.34253 

.423 

.36488 

.310 

.28880 

.348 

.31739 

.386 

.34317 

.424 

.36541 

.311 

.28958 

.349 

.31811 

.387 

.34380 

.425 

.36594 

.312 

.29036 

.350 

.31882 

.388 

.34444 

.426 

.36646 

.313 

.29115 

.351 

.31954 

.389 

.34507 

.427 

.36698 

.314 

.29192 

.352 

.32025 

.390 

.34569 

.428 

.36750 

.315 

.29270 

.353 

.32096 

.391 

.34632 

.429 

.36802 

.316 

.29348 

.354 

.32167 

.392 

.34694 

.430 

.36853 

.317 

.29425 

.355 

.32237 

.393 

.34756 

.431 

.36904 

.318 

.29502 

.356 

.32307 

.394 

.34818 

.432 

.36954 

.319 

.29580 

.357 

.32377 

.395 

.34879 

.433 

.37005 

.320 

.29656 

.358 

.32447 

.396 

•3494p 

.434 

.37054 

.321 

.29733 

.35Q 

.32517 

.397 

.35001 

.435 

.37104 

.322 

.29810 

.360 

.32587 

.398 

.35062 

.436 

.37153 

142        MENSURATION    OF   AREAS,  LINES,  AND    SURFACES. 


Tabh 

i  of  ike  Zones  of  a  Circle — Continued. 

Height.]   Area. 

Height. 

Area. 

Height. 

Area.    Height. 

Area. 

.437 

.37202 

.453 

.37931 

.469 

.38549 

.485 

.39026 

.438 

.37250 

.454 

.37973 

.470 

.38583 

.486 

.39050 

.439 

.37298 

.455 

.38014 

.471 

.38617 

.487 

.39073 

.440 

.37346 

.456 

.38056 

.472 

.38650 

.488 

.39095 

.441 

.37393 

.457 

'.38096 

.473 

.38683 

.489 

.39117 

.442 

.37440 

.458 

.38137 

.474 

.38715 

.490 

.39137 

.443 

.37487 

.459 

.38177 

.475 

.38747 

.491 

..39156 

.444 

.37533 

.460 

.38216 

.476 

.38778 

.492 

.39175 

.445 

.37579 

.461 

.38255 

.477 

.38808 

.493 

.39192 

.446 

.37624 

.462 

.38294 

.478 

.38838 

.494 

.39208 

.447 

.37669 

.463 

.38332 

.479 

.38867 

.495 

.39223 

.448 

.37714 

.464 

.38369 

.480 

.38895 

.496 

.39236 

.449 

.37758 

.465 

.38406 

.481 

.38923 

.497 

.39248 

.450 

.37802 

.466 

.38443 

.482 

.38950 

.498 

.39258 

.451 

.37845 

.467 

.38479 

.483 

.38976 

.499 

.39266 

.452 

.37888 

.468 

.38514 

.484 

.39001 

.500 

.39270 

To  find  the  Area  of  a  Zone  by  the  above  Table. 

Rule  1. —  When  the  zone  is  less  than  a  semicircle,  divide  the 
height  by  the  diameter,  and  find  the  quotient  in  the  column 
of  heights.  Take  out  the  area  opposite  to  it  in  the  next  col- 
umn on  the  right  hand,  and  multiply  it  by  the  square  of  the 
longest  chord  ;  the  product  will  be  the  area  of  the  zone. 

Example. — Required  the  area  of  a  zone,  the  diameter  of 
which  is  50,  and  its  height  15  ? 

15^50  =  . 300;  and  .300,  as  per  table,  =.28088. 
Hence,     .28088  x  502= 702.2,  the  area  of  the  zone. 

Rule  2. —  When  the  zone  is  greater  than  a  semicircle,  take  the 
height  on  each  side  of  the  diameter  of  the  circle,  and  find,  by 
Rule  1,  their  respective  areas;  add  the  areas  of  these  two 
portions  together,  and  the  sum  will  be  the  area  of  the  zone. 

Example. — Required  the  area  of  a  zone,  the  diameter  of 
the  circle  being  50,  and  the  height  of  the  zone  on  each  side  of 
the  diameter  of  the  circle  20  and  15  respectively. 


MENSURATION    OF   AREAS,  LINES,  AND    SURFACES.         143 

20-r-50=.400;  A00,  as  per  table,  =  .35182 ;  and  .35182  x 
502=879.55. 

15-^50=.300;  .300,  asper  table,  =.28088  ;  and  .28088  x 
502= 702/2. 

Hence,     879.55  +  702.2  =  1581.65,  the  result  required. 

Note. — -When,  in  the  division  of  a  height  by  the  chord,  the 
quotient  has  a  remainder  after  the  third  place  of  decimals,  and 
great  accuracy  is  required, 

Take  the  area  for  the  first  three  figures,  subtract  it  from 
the  next  following  area,  multiply  the  remainder  by  the  said 
fraction,  and  add  the  product  to  the  first  area ;  the  sum  will 
be  the  area  for  the  whole  quotient. 

Example. — What  is  the  area  of  a  zone  of  a  circle,  the 
greater  chord  being  100  feet,  and  the  breadth  of  it  14  feet  3 
inches  % 

14  feet  3  inches=U.25,  and  14.25  -f- 100  =  .1425  ;  the  tab- 
ular- length  for  .142  =  .14007 ',  and  for  .143=. 14103,  the  differ- 
ence  between  which  is  .00096. 

Then,  .5  x  .00096  =  .000480. 

Hence,  .142   =.14007,'     . 

.0005 =.00048 

.14055,  the  sum  by  which  the  square 
of  the  greater  chord  is  to  be  multiplied;  and  .14055  xl002=: 
1405.5  feet. 


^ 


144        MENSURATION   OF   AREAS,  LINES,  AND    SURFACES. 


PEOMISCUOUS  EXAMPLES. 

1.  If  a  load  of  wood  is  8  feet  long,  3  feet  10  inches  wide, 
and  6  feet  6  inches  high,  what  are  its  contents  ? 

Ans.  1.72  cords. 

2.  Add  |  of  a  ton  to  -j^  of  a  cwt.  Ans.  12.329  cwt. 

3.  What  are  the  contents  of  a  board  25  feet  long  and  3  feet 
wide?  Ans.  75  feet. 

4.  What  is  the  difference  between  the  contents  of  two  floors ; 
one  being  37  feet  long  and  27  feet  wide,  and  the  other  40  feet 
long  and  20  feet  wide?  Ans.  199  feet. 

5.  How  many  yards  of  paper  that  is  30  inches  wide  will  it 
require  to  cover  the  wall  of  a  room  that  is  15^  feet  long,  11^ 
feet  wide,  and  7  J  feet  high  ?  Ans.  55.2833  yards. 

6.  If  i  of  a  post  stands  in  the  mud,  J  in  the  water,  and  10 
feet  above  the  water,  what  is  the  length  of  the  post  ? 

Ans.  18.182  feet. 

7.  What  fraction  is  -that  to  which  if  f-  of  -J  be  added  the 
sum  will  be  1  ?  Ans.  •§-§• 

8.  If  the  earth  make  one  complete  revolution  in  23  hours 
5£  minutes  3  seconds,  in  what  time  does  it  move  one  degree  ? 

Ans.  3  mm.  59.3417  seconds. 

9.  From  a  plank  26  inches  broad  a  square  yard  and  a  half 
is  to  be  sawed  off;  what  distance  from  the  end  must  the  line 
be  struck  ?  Ans.  6. 23  feet. 

10.  What  is  the  side  of  a  triangle  that  may  be  inscribed  in 
a  circle,  the  circumference  of  which  is  1000  feet? 

Ans.  275.6556  feet. 

11.  How  large  a  square  field  can  be  made  in  a  circle  of 
100  rods  in  diameter? 

Ans.  22  rods  2  yards  2  feet  4.5  inches. 

12.  A  rectangular  field  is  12  rods  2  yards  2  feet  and  3 
inches  in  length,  by  9  rods  and  1  yard  in  breadth ;  what  is 
its  area  in  square  yards?  Ans.  3471.875  yards.  . 


MENSURATION    OF  AREAS,  LINES,  AND    SURFACES.        145 

13.  The  sides,  of  a  triangular  plot  of  ground  are  24,  36, 
and  48  feet ;  what  is  its  area  in  square  feet  ? 

Ans.  418.282  feet. 

14.  In  turning  a  chaise  within  a  ring  of  a  certain  diameter, 
the  outer  wheel  made  two  turns  while  the  inner  wheel  made 
but  one,  the  wheels  being  four  feet  in  diameter,  and  five  feet 
asunder  on  the  axle-tree ;  what  was  the  circumference  of  the 
track  described  by  the  outer  wheel?  Ans.  62.832  feet. 

15.  The  ball  on  the  top  of  a  church  is  6  feet  in  diameter ; 
what  did  the  gilding  of  it  cost  at  8  cents  per  square  inch  ? 

Ans.  1302.884  dollars. 

16.  A  roof  of  a  house  is  24  feet  8  inches  by  14.5  feet,  and 
is  to  be  covered  with  lead  weighing  8  lbs.  per  foot;  what 
will  be  the  weight  of  the  lead  required  ? 

Ans.  2861.324  lbs. 

17.  The  area  of  an  equilateral  triangle,  whose  base  falls  on 
the  diameter,  and  its  vertex  in  the  middle  of  the  arc  of  a  semi- 
circle, is  equal  to  100 ;  what  is  the  diameter  of  the  semicircle? 

Ans.  26.32148. 

18.  The  distance  of  the  centres  of  two  circles,  the  diameters 
of  which  are  each  50,  is  equal  to  30  j  what  is  the  area  of  the 
space  inclosed  between  the  two  circles  by  arcs  of  their  circum- 
ferences? Ans.  559.115.* 

19.  In  the  latitude  of  London,  the  distance  around  the 
earth,  measured  on  that  parallel,  is  about  15,550  miles ;  now, 
as  the  earth  revolves  in  23  hours  and  56  minutes,  at  what 
rate  per  hour  does  the  city  of  London  move  from  west  to 
east?  Ans.  649.7214  miles  per  hour. 

20.  A  father  left  his  son  an  estate,  ^  of  which  he  ran 
through  in  8  months ;  f-  of  the  remainder  lasted  him  12  months 
longer,  wheit  he  had  barely  $820  left ;  what  sum  did  his  fa- 
ther leave  him  ?  Ans.  $1913.34. 

21.  There  is  a  segment  of  a  circle  the  chord  of  which  is  60 
feet,  its  versed  sine  10  feet ;  what  will  be  the  versed  sine  of 
that  segment  of  the  same  circle  when  the  chord  is  90  feet? 

Ans.  28.2055. 
*  By  Table  of  Areas  of  the  Segments  of  a  Circle,  p.  134. 
G 


146        MENSURATION    OF    AREAS,  LINES,  AND    SURFACES. 

22.  If  a  line  144  feet  long  will  reach  from  the  top  of  a 
fort  to  a  point  on  the  opposite  side  of  a  river  that  is  64  feet 
wide,  what  is  the  height  of  the  fort  above  that  point  1 

Ans.  128.99  feet. 

23.  A  certain  room  is  20  feet  long,  16  feet  wide,  and  12  feet 
high ;  how  long  must  a  line  be  to  extend  from  one  of  the  lower 
corners  to  an  opposite  upper  corner'?        Ans.  28.2843  feet. 

24.  Two  ships  sail  from  the  same  por.t ;  one  sails  due  north 
128  miles,  the  other  due  east  72  miles;  how  far  are  the  ships 
from  each  other?  Ans.  146.86+  miles. 

25.  There  are  two  columns  in  the  ruins  of  Persepolis  left 
standing  upright ;  one  is  70  feet  above  the  plain,  and  the  oth- 
er 50 ;  in  a  straight  line  between  these  stands  an  ancient 
statue  5  feet  in  height,  the  head  of  which  is  100  feet  from  the 
summit  of  the  higher,  and  80  feet  from  the  top  of  the  lower 
column ;  required  the  distance  between  the  tops  of  the  two 
columns'?  *  Ans.  143.543  feet. 

26.  The  height  of  a  tree  growing  in  the  centre  of  a  circular 
island  100  feet  in  diameter  is  160  feet,  and  a  line  extending 
from  the  top  of  it  to  the  further  shore  is  400  feet ;  what  is 
the  breadth  of  the  stream,  assuming  the  land  on  each  side  of 
the  water  to  be  level  %  Ans.  316.6065  feet. 

27.  A  ladder  70  feet  long  is  so  placed  as  to  reach  a  win- 
dow 40  feet  from  the  ground  on  one  side  of  a  street,  and  with- 
out removing  it  at  the  foot,  will  reach  a  window  30  feet  high 
on  the  other  side  ;  what  is  the  breadth  of  the  street  ? 

Ans.  120.6911/^. 

28.  If  a  tree  stand  on  a  horizontal  plane  80  feet  in  height, 
at  what  height  from  the  ground  must  it  be  cut  off  so  that  the 
top  of  it  may  fall  on  a  point  40  feet  from  the  bottom  of  the 
tree,  the  end  where  it  was  cut  off  resting  on  the%tump  ? 

Ans.  30  feet. 

29.  Four  men,  A,  B,  C,  D,  bought  a  grindstone,  the  diam- 
eter of  which  was  4  feet ;  they  agreed  that  A  should  grind 
off  his  share  first,  and  tnat  each  man  should  have  it  alternately 
until  he  had  worn  off  his  share  ;  how  much  will  each  man  grind 
off?         Ans.  A  3.215  +  ,  B  3.81  +  ,  C  4.97  +  ,  D  12  inches. 


MENSURATION    OF    AREAS,  LINES,  AND    SURFACES.        147 

30.  The  classification  of  a  school  is  as  follows,  viz.,  -^  of 
the  boys  are  taught  geometry,  -|  grammar,  -^5-  arithmetic,  ^ 
writing,  and  9  reading;  what  is  the  number  in  each  branch'? 

j,        (5  geometry,  30  grammar,  24  arithmetic, 
'  \  12  writing,  and  9  reading. 

31.  A  certain  general  has  an  army  of  141,376  men.  How 
many  must  he  place  in  rank  and  file  to  form  them  into  a 
square?  Ans.  376. 

32.  If  *he  area  of  a  circle  be  1760  yards,  how  many  feet 
must  the  side  of  a  square  measure  to  contain  that  quantity  ? 

Ans.  125.8571  jeet. 

33.  If  the  diameter  of  a  round  stick  of  timber  be  24  inches, 
how  large  a  square  stick  may  be  hewn  from  it  ? 

Ans.  16.97  inches. 

34.  To  set  out  an  orchard  of  2400  mulberry  trees  so  that 
the  length  shall  be  to  the  breadth  as  3  to  2,  and  the  distance 
of  each  tree  one  from  the  other  7  yards,  how  many  trees  must 
there  be  in  the  length  of  the  orchard,  and  how  many  in  its 
breadth,  and  how  many  square  yards  of  ground  do  they  stand 
on  ?  C  Trees  in  length,  60. 

Ans.  <  Trees  in  breadth,  40. 

{Square  yards,     1 1 7,600. 

35.  Suppose  the  expense  of  paving  a  semicircular  plot  of 
ground,  at  30  cents  per  square  foot,  amounted  to  $25.63, 
what  is  the  diameter  of  it1?  Ans.  14.75  feet. 

36.  Two  sides  of  an  obtuse-angled  triangle  are  20  and 
40  poles ;  what  must  be  the  length  of  the  third  side,  that 
the  triangle  may  contaiiFJust  an  acre  ? 

Ans.  58.876  ^ofes. 
37-  If  two  sides  of  an  obtuse-angled  triangle,  the  area  of 
which  is =60  x  -/3,  are  12  and  20,  what  is  the  third  side? 

Ans.  28. 
38.  If  an  area  of  63  feet  is  cut  off  from  a  triangle,  the 
three  sides  of  which  are  13,  20,  and  21  feet,  by  a  line  paral- 
lel to  the  longest  side  or  base  of  the  triangle,  what  are  the 
lengths  of  the  sides  of  the  triangle  which  will  include  that 
area? 


148        MENSURATION    OF    AREAS,  LINES,  AND    SURFACES. 

Operation. — A  triangle  of  the  above  dimensions  has  an  area  of  126. 
See  Rule,  p.  51. 

Then,  as  126  {—area  of  triangle) :  J-f^  (  —  area  of  required  triangle)'.'. 
hyp.2  (400) :  hyp/2  (200)=square  of  hyp.  of  required  triangle;  and  -y/200 
=  14.142  =square  root  of  square  of  hyp.— hyp.'  of  required  triangle. 

Hence,  as  hyp.  (20) :  base  (21) : :  hyp.'  (14.142) :  base  of  required  trian- 
gle (14.849). 

Consequently,  14.849  is  the  base,  14.142  is  the  hyp.,  and  by  Rule,  p. 
53,  V  14.1422  -14.8492 =4.527,  the  length  of  the  remaining  side. 

39.  Seven  men  bought  a  grindstone  of  60  inches  in  diam- 
eter, each  paying  \  part  of  the  cost;,  what  part  of  the  diam- 
eter can  each  grind  down  for  his  share  ? 

(The  1st,  4.4508  ;  2d,  4.8400 ;  3d,  5.3535 ; 
Ans.  <         4th,  6.0765  ;  5th,  7.2079 ;  6th,  9.3935  ; 
(  •     and  the  7th,  22.6778. 


This  problem  may  be  thus  constructed,  Fig.  69  : 

On  the  radius,  a  c,  describe  a  semicircle ;  also  divide  a  c  into  as  many 
equal  parts,  c  d,  d  e,  ef  &c,  as  there  are  shares,  and  erect  the  perpen- 
diculars d  I,  e  m,fn,  &c,  meeting  the  semicircle  described  on  a  c  in  /, 
m,  n,  o,  p,  q. 

Then,  with  the  centre  c  and  radii  c  I,  c  in,  c  n,  &c,  describe 
circles,  and  the  diameter  which  each  is  to  grind  down  will  be  thus 
shown. 


Fig.  69. 


For,  the  square  of  the  chords  or  radii  c  I,  cm,  en,  &c,  arc  as  the  co- 
sines c  d,c  e,  cf  <fec. 

40.  A  gentleman  has  a  garden  100  feet  long  and  80  feet 
broad,  and  a  gravel  walk  is  to  be  made  of  an  equal  width 


MENSURATION   OF  AREAS,  LINES,  AND    SURFACES.         149 

half  around  it ;  what  must  be  the  width  of  the  walk  so  that 
it  will  take  up  just  half  the  gr.ound  ?* 

Operation. — 100  X  80= 8000 =area  of  garden. 

100  +  80 

=90=half  length  of  half  the  sides  of  garden. 

2 
8000 
902 — - =4100=araz  of  walk  if  the  garden  was  a  square. 

Hence,  V 4100  =64.0312,  and  64.0312  -  90=25.9688=  width  of  the 
walk. 

41.  In  the  midst  of  a  meadow,  well  covered  with  grass, 
It  just  took  an  acre  to  tether  an  ass ; 

How  long  was  a  line,  that,  reaching  all  round, 
Restricted  his  grazing  to  an  acre  of  ground  ? 

Ans.  39.2507  yards. 

42.  A  maltster  has  a  kiln  that  is  16  feet  6  inches  square; 
it  is  necessary  to  pull  it  down,  and  build  a  new  one  that  will 
dry  three  times  as  much  at  a  time  as  the  old  one  did ;  what 
must  be  the  length  of  its  side  %  Ans.  28.58  feet. 

43.  In  a  round  garden  containing  75  square  rods,  how 
large  a  square  garden  can  be  laid  out  1 

Ans.  47.7475  square  rods. 

44.  If  a  circular  garden  contain  75  square  rods,  what 
must  be  the  side  of  a  square  field  that  would  inclose  it  ? 

Ans.  9.772  rods. 

45.  There  is  a  circular  field  25  rods  in  diameter ;  what  is 
the  difference  of  the  areas  of  the  inscribed  and  circumscribed 
squares,  and  how  much  do  they  differ  from  the  areas  of  the 
field? 

f312.5  rods,  the  difference  of  the  squares;  134.1261 
Ans.  •<      rods,  the  circumscribed  square  more  than  the  area; 
C     178.374  rods,  inscribed  square  less  than  the  area. 


*  This  problem  may  be  constructed  thus: 
Let  abed  represent  the  garden  ;  make  c  e  = 
c  b,  and  with  the  centre  d  and  radius  d  e  de- 
scribe the  semicircle  g  ef.  Make  b  i=$b  g, 
b  l=\bf,  and  complete  the  rectangle  i  b  I  h, 
and  the  result  is  obtained. 


Fig.  70. 

SCAe 


150        MENSURATION   OF  AREAS,  LINES,  AND    SURFACES. 

46.  Two  persons  start  from  the  same  place ;  one  goes  south 
4  miles  per  hour,  the  other  west  5  miles  per  hour;  how  far 
apart  are  they  in  9  hours?  A ns.  57.6281  miles. 

47.  The  four  sides  of  a  field,  Fig.  71,  the  diagonals  of  which 
are  equal  to  each  other,  are  25,  35,  31,  and  19  poles;  what 
is  its  area  ? 

Construction. — In  this  question,  the  sums  of  the  squares  of  the  oppo- 
site sides  of  the  trapezium  being  equal  (d  c2  (19)  -\-a  b2  (35)  =  1586,  and 
d  a2  (25) +c  b2  (31)  =  1586),  the  figure  may  be  constructed  as  follows : 

Fig.  71. 


Draw  a  b  and  a  e  at  right  angles,  and  each  equal  to  the  longest  of  the 
given  sides  (35) ;  join  b  e,  and  from  the  points  e  and  b,  with  radii  equal 
to  the  two  remaining  opposite  sides  (25  and  31)  respectively,  describe 
arcs  intersecting  in  c  on  the  farther  side  of  b  e;  join  a  c,  and  draw  bf 
at  right  angles  to  it.  With  the  centre  c,  and  radius  equal  to  the  re- 
maining side  (19),  describe  an  arc  cutting  6/produced  in  d.  Join  a  d 
and  c  d;  then  will  a  b  c  d  be  the  figure  required. 

Hence,  in  the  two  triangles  a  b  d  and  e  a  c,  we  have  the  two  sides  b  a, 
a  d  equal  the  two  a  e,  e  c,  each  to  each ;  and  the  angles  a  b  d  and  e  a  c 
equal  (each  being  the  complement  of  b  a/),  and  e  c  and  a  d  similarly 
situated;  wherefore  b  d—a  c. 


Calculation. — On  b  e  let  fall  the  perpendiculars  c  g  and  a  h. 
Now  b  e*=a  b2+a  e2=352X2  ;  b  e  =  -/352x2=35\/2=49.4975. 
a  h=bh=\b  e=24.7487. 

,,    c         ,  Ka   ,        bc2+be2-ce>     312+2~x35"2-25a       2 

By  formula,  p.  56,  b  g= 

:28.1428. 


26  e  2x49.4975  98.995 

g  h=bg-b  A=28.1428-24.7487=3.3941. 

c^  =  -/6c2-65r3=-/312-28.14282  =  -/ 168.9828=12.9993. 

By  similar  triangles  (see  Note,  p.  50),  a  h+c  g  (37.748)  :g  h  (3.3941) 
: : a  h  (24.7487);  h  i=2.2253. 

ai  =  Va  h2  +  h  i2  =  -/24.7487a  +  2.22532  =  V 612.4981  +  4.9520  = 
V  617.4501  =24:8485. 


MENSURATION    OP  AREAS,  LINES,  AND    SURFACES.        151 

Again,  by  similar  triangles,  h  i  (2.2253)  :  h  g  (3.3941):  \a  i  (24.8485) 

:  a  c= 37.8997 =6  d ;  now,  by  Rule,  p.  57,  the  area  of  the  trapezium*  abed 

acxbf+fd    acXbd    ac2     37.89972  1QQ_       .  : 

= si — ^—  = =—-= —  718.1936  poles=4:  ac.  1  ro. 

2  2  2  2  r 

38  po.  5  yds.  7.7076 feet,  Ans. 

48.  A  messenger  traveling  8  miles  an  hour  was  sent  to 
Mexico  with  dispatches  for  the  army;  after  he  had  gone  51 
miles,  another  was  sent  with  countermanding  orders  who 
could  go  19  miles  as  quick  as  the  former  could  16  ;  how  long 
will  it  take  the  latter  to  overtake  the  former,  and  how  far 
must  he  travel  ? 

Operation. — If  the  first  messenger  travels  8  miles  in  an  hour,  and  the 
second  19  while  the  first  travels  16,  the  second  travels  1.5  miles  an  hour 


Pr-*) 


faster  than  the  first. 


Then,  51-f-1.5=34  hours,  and  3ixd. 5 =323  =number  of  miles  traveled 
by  second  messenger  when  he  overtook  the  first. 

51 

Hence,  34+— -=40.375,  and 40.375  X  8 =323 miles  =  ihe  distance  reach- 
8  • 

ed  by  first  messenger. 

49.  The  hour  and  minute  hands  of  a  watch  are  exactly  to- 
gether at  12  o'clock ;  when  are  they  next  together? 

Operation. — The  velocities  of  the  two  hands  of  a  watch  are  to  each 
other  as  12  to  1 ;  therefore  the  difference  of  velocities  is  12  —  1  =  11. 

Then, as  11  :  i  "  *  \  I  : : :  1 :  i  J »■  *> ""  2?ft«C-  ?*** 

\  12  X  2  )  \2h.l0m.  54-jSj-  sec,  2d  tune. 

50.  A  person  being  asked  the  hour  of  the  day,  replied, 
The  time  past  noon  is  equal  to  f  of  the  time  till  midnight ; 
what  was  the  time?  Ans.  20  minutes  past  5. 

Operation.— If  the  time  required  is  |  of  the  time  to  midnight,  then  the 
whole  time  from  noon  to  midnight  (12  hours  or  720  minutes)  is  divided 
into-fd+f+1). 

Hence,  if  -|=720,  £=80  minutes,  and  ^=320  minutes,  or  5  hours  and 
20  minutes,  the  Ans. 

51.  A  person  being  asked  the  time  of  day,  replied  that  | 
of  the  time  from  noon  was  equal  to  -^T  of  the  time  to  mid- 
night; what  was  the  time?  Ans.  40  minutes  past  4. 

*  a  c  and  b  d  being  equal. 


152        MENSURATION    OF   AREAS,  LINES,  AND    SURFACES. 

52.  What  is  the  radius  of  a  circular  acre? 

Operation. — Side. of  a  square  (p.  81)  X  1.128= diameter  of  an  equal 
circle. 

The  side  of  a  square  acre  (p.  13)  is  208.710321,  which,  X  1.128  = 
235.5,  and  235.5^2  to  obtain  radius=ll7. 75  feet. 

53.  The  time  of  the  day  is  between  4  and  5,  and  the  hour 
and  minute  hands  are  exactly  together;  what  is  the  time? 

Operation. — The  speed  of  the  hands  is  as  1  to  11. 
4  hours  X  60=240,  and  240-f-ll=21  min.  and  49^  sec,  which,  added 
to  4,  =4  hours  21  min.  and  49^-  sec. 

54.  A  person  being  asked  what  o'clock  it  was,  replied  that 
it  was  between  5  and  6  ;  but,  to  be  more  particular,  the  min- 
ute-hand was  as  far  beyond  the  6  as  the  hour-hand  wanted  of 
being  to  the  6  ;  that  is,  that  the  hour  and  minute  hands  made 
equal  acute  angles  with  a  line  passing  from  the  12  through 
the  6  ;  required  the  time. 

Operation. — 5  hours— 300  minutes,  and  6  hours = 360  minutes. 

Then,  300  4- the  time  by  the  hour-hand  past  5=360— the  time  by  the 
minute-hand  past  6. 

As  the  relative  speed  of  the  hour  and  minute  hands  is  as  1  to  12, 
300  +  1,=360-12. 

Q£f) 300 

Consequently,  ~\%—^i  36^§=the  space  between  the  hour 

1  ~rl2 

and  minute  hands,  which,  -J- 2  to  obtain  the  half  space  (each  side  of  the 
6),  gives  2  min.  18-^g-  sec,  which,  added  to  5  hours  and  30  min.,  =5  hours 
32  min.  and  18^-  sec,  the  time  required. 

4 

55.  Two  persons,  A  and  B,  start  at  the  same  time  to  meet 
each  other  when  apart  100  miles ;  after  7  hours  they  meet, 
when  it  appears  that  A  had  ridden  \\  miles  per  hour  faster 
than  B ;  at  what  rate  per  hour  did  each  ride  % 

Ans.  A  7.893,  B  6.392  miles  per  hour. 

56.  Swift  can  travel  7  miles  in  -§  of  an  hour,  but  Slow 
can  travel  only  5  miles  in  -^  of  an  hour ;  »both  started  from 
one  point  at  the  same  time  to  walk  a  distance  of  12  miles; 
how  much  sooner  will  Swift  arrive  than  Slow  ? 

Ans.  12.467  seconds. 


MENSURATION   OP  AREAS,  LINES,  AND    SURFACES.        153 

57.  At  a  certain  time  between  two  and  three  o'clock,  the 
minute-hand  of  a  clock  was  between  three  and  four ;  within 
an  hour  after,  the  hour  and  minute  hands  had  exactly  changed 
places  with  each  other ;  what  was  the  precise  time  when  the 
hands  were  in  the  first  position  ? 

Ans.  2  h.  15  m.  56^-  sec. 

58.  If  a  traveler  were  to  leave  New  Haven  at  8  o'clock  on 
a  morning,  and  walk  toward  Albany  at  the  rate  of  3  miles  an 
hour,  and  another  traveler  were  to  set  out  from  Albany  at 
4  o'clock  in  the  evening,  and  walk  toward  New  Haven  at  the 
rate  of  4  miles  an  hour,  whereabout  on  the  road  would  they 
meet,  supposing  the  distance  to  be  130  miles  ? 

Ans.  69.4286  miles  from  New  Haven. 

59.  A  thief,  escaping  from  an  officer,  has  40  miles  the 
start,  and  travels  at  the  rate  of  5  miles  an  hour ;  the  officer 
in  pursuit  travels  at  the  rate  of  7  miles  an  hour ;  how  far 
must  he  travel  before  he  overtakes  the  thief? 

Ans.  20  hours,  and  140  miles. 

60.  If  12  oxen  graze  3^  acres  of  grass  in  4  weeks,  and  21 
oxen  10  acres^in  9  weeks,  how  many  oxen  would  it  require  to 
graze  24  acres  in  18  weeks,  the  grass  to  be  growing  ?• 

Operation. — Each  ox  grazes  a  certain  quantity  in  each  week,  which 
we  suppose  to  be  100  pounds,  and  of  the  whole  quantify  grazed  in  each 
case,  a  part  must  have  grown  during  the  time  of  grazing. 

Then,  by  the  first  condition, 

1 2  X  4  X  100—4800  lbs.  =  whole  quantity  on  3£  acres  for  4  weeks. 
4800-7-3^=1410  lbs.=whole  quantity  on  1  acre  for  4  weeks. 
By  the  second  condition, 

21 X  9  X  100=18900  lbs.  =whole  quantity  on  10  acres  for  9  weeks. 
18900-7-10  =  1890  lbs.=whole  quantity  on  1  acre  for  9  weeks. 
1890  —  1440=450  lbs.  =  the  quantity  grown  on  an  acre for  9— 4= 5  weeks. 
450-f-9—4=90  lbs.  =  the  quantity  which  grows  on  each  acre  for  1  week. 
90x3^x4  =  1200  lbs.  =  quantity  grown  on  3^  acres  for  4  weeks. 
4800—1200=3600  lbs. = original  quantity  of  grass  on  3^  acres. 
3600-7-3^=1080  lbs. ^original  quantity  on  1  acre. 
And  by  the  last  condition, 

24  X  1080=25920  lbs.  ^original  quantity  on  24  acres. 
24  X  90  X  18  =  38880  lbs.  =quantity  which  grows  on  24  acres  in  18  weeks. 
G2 


154        MENSURATION    OF   AREAS,  LINES,  AND    SURFACES. 

25920+38880=64800  lbs.=whole  quantity  on  24  acres  for  18  weeks. 
64800-r-18=3600  lbs.— quantity  to  be  grazed  from  24  acres  each  week. 
3600-r-100  =36  =m«?i&er  of  oxen  required  to  graze  the  whole. 

61.  A  tract  of  land,  exactly  square,  is  inclosed  by  a  three- 
railed  fence  ;  the  length  of  each  rail  is  15  feet,  and  the  num- 
ber of  rails  in  the  fence  is  equal  to  the  number  of  acres  in- 
closed ;  required  the  area  of  this  tract  in  acres,  and  the  length 
of  its  side  in  feet. 

Operation. — If  the  tract  of  land  was  inclosed  by  one  rail,  then,  15-r- 
(4x3) =1.2  5  feet,  the  length  of  its  side. 

Then,  if  43,560  square  feet  make  an  acre,  as  1.252 :  43,560: :1  rail: 
27878.4,  the  number  of  rails  in  the  fence,  or  the  number  of  acres  in  the 
tract;  and  (27878.4 X  15)-r-(4 X 3)  =34,848,  the  length  of  the  side  in 
feet. 

62.  What  is  the  radius  of  a  circular  acre  % 

Operation. — Side  of  a  square  X  1.128  =  diameter  of  an  equal  circle. 

By  table,  p.  13,  208.710321  =the  side  of  a  square  acre. 

Then,  208.710321x1.128=235.50,  which,  -r-2  (for  radius),  =117.75 

feet, 

63.  There  is  an  island  20  miles  in  circumference,  and  three 
men  start  together  to  travel  the  same  way  about  it ;  A  goes 
2  miles  per  hour,  B  4  miles  per  hour,  and  C  6  miles  per  hour ; 
in  what  time  will  they  come  together  again  ? 

Ans.  10  hours. 

64.  A  hare  starts  12  rods  before  a  hound,  but  is  not  per- 
ceived by  him  till  she  has  been  off  1^  minutes ;  she  runs  at 
the  rate  of  36  rods  a  minute,  and  the  dog,  on  view  of  her, 
makes  after  her  at  the  rate  of  40  rods  a  minute ;  how  long 
will  the  course  hold,  and  what  distance  will  the  dog  run  ? 

Ans.  14 J  minutes,  and  he  will  run  570  rods. 


MENSURATION    OF    SOLIDS. 


155 


MENSURATION  OF  SOLIDS. 

OP    CUBES     AND     PAEALLELOPIPEDONS. 

Cube. 
Definition.     A  solid  contained  by  six  equal  square  sides. 

%       To  ascertain  the  Contents  of  a  Cube,  Fig.  72 
Eule. — Multiply  a  side  of  the  cube  by  itself,  and  that  prod- 
uct again  by  a  side,  and  this  last  product  will  give  the  con- 
tents required. 

Or,  s^S,  s  representing  the  length  of  a  side,  and  S  the  con- 
tents. 

Fig.  72. 


X 

~~[\ 

m 

"\ 

ap 

Example. — The  side  a  b  of  the  cube,  Fig.  72,  is  12  inches; 
what  are  the  contents  of  it  f 

12.X  12  X  12  =  1728  inches. 

Ex.  2.  The  side  of  a  cube  is  15  inches ;  what  are  its  con- 
tents in  feet  and  inches  % 

^725.1.953125/^, or  1  foot  and  H^Voo  inches. 
Ex.  3.  The  sides  of  a  cube  are  12.5  feet;  what  are  its 
contents  in  cubic  feet  and  yards  ? 

A       C1953.125  cubic  feet. 
t     72.338  cubic  yards. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 


156 


MENSURATION   OF    SOLIDS. 


Parallelopipedon. 

Definition.     A  solid  contained  by  six  quadrilateral  sides,  every 
opposite  two  of  which  are  equal  and  parallel. 

To  ascertain  the  Contents  of  a  Parallelopipedon,  Fig.  73. 
Rule. — Multiply  the  length  by  the  breadth,  and  that  prod- 
uct again  by  the  depth,  and  this  last  product  will  give  the  9 
contents  required. 
Or,  lxbxd=S. 

Fig.  73. 


Example. — The  length  a  b,  Fig.  73,  is  15,  the  breadth  c  d 
12,  and  the  depth  c  b  is  11  inches;  what  are  the  contents? 
15  X  12  X  11  =  1980  inches. 

Ex.  2.  The  length  of  a  parallelopipedon  is  15  feet,  and 
each  side  of  it  is  21  inches ;  what  are  its  contents? 

Ans.  45.9375  feet. 

Ex.  3.  The  dimensions  of  a  parallelopipedon  are  20  feet  in 
length,  11.5  in  breadth,  and  7  in  depth ;  what  are  its  contents 
in  feet?  Ans.  1610  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 


PEISMS,  PRISMOIDS,  AND   WEDGES. 

Prisms. 

Definition.  Solids,  the  ends  of  which  are  equal,  similar,  and 
parallel  planes,  and  the  sides  of  which  are  parallelograms. 

Note. — When  the  ends  of  a  prism  are  triangles,  it  is  called  a  trian- 
gular prism ;  when  rhomboids,  a  rhomboidal  prism ;  when  squares,  a 
square  prism;  when  rectangles,  a  rectangular  prism,  &c. 


MENSURATION   OP   SOLIDS. 


157 


To  ascertain  the  Contents  of  a  Prism,  Figs.  74  and  75. 
Rule. — Multiply  the  area  of  the  base  d  efby  the  height, 
and  the  product  will  give  the  contents  required. 
Or,  axh=S.  r 

Fig.  74.  Fig.  75. 


Example. — A  triangular  prism,  ahcdef  Fig. 75,  has  sides 
of  2.5  feet,  and  a  length  c  d  of  10  feet ;  what  are  its  contents  ? 

(By  Eule,  p.  52)  -^-=1.5625,  which,  X  1.732^2.70625  =area  of  end 
a  be,  and  2.70625  X 10  =  27.0625  =feet. 

Ex.  2.  A  side  of  the  end  of  a  triangular  prism  is  18  inches, 
and  the  length  of  the  prism  is  9  feet ;  what  are  its  contents  ! 

Ans.  8.7 682  feet. 

Ex.  3.  What  is  the  solidity  of  a  prism  15  feet  in  length, 
the  ends  of  which  are  hexagonal,  with  sides  16  inches  in 
length  f 

(By  Eule,  p.  60)  162X  2.5981  =  665.1136 = square  of  a  side  multiplied 
by  the  tabular  number  for  the  area  of  a  hexagon;  then,  665.1136x15  = 
91)76.704,  the  contents  required. 

Ex.  4.  The  sides  of  an  octagonal  prism  are  3  feet,  and  its 
height  6.75  feet;  what  are  its  contents  in  feet  and  yards? 

*  A        ("293.3388/^. 
t   10.8644  yards. 

"  Centre  of  Gravity.     For  rule,  see  Mensuration  of  Areas, 
Lines,  and  Surfaces,  p.  100.  «. 


158 


MENSURATION    OF   SOLIDS. 


Prismoids. 

Definition.  Figures  alike  to  a  prism,  but  having  only  one  pair 
of  their  sides  parallel. 

Note. — Prismoids,  alike  to  prisms,  derive  their  designation  from  the 
figure  of  their  ends,  as  triangular,  square,  rectangular,  pentagonal,  &c. 

To  ascertain  the  Contents  of  a  Prismoid,  Fig.  76. 

Rule. — To  the  sum  of  the  areas  of  the  two  ends,  abed, 
efgh,  add  four  times  the  area  of  the  middle  section  at  i  k 
parallel  to  them ;  multiply  this  sum  by  -J-  of  the  height,  and 
the  product  will  give  the  contents  required.* 

Or,  a-\-a/-\-4mxh-r-6  =  Si  a  and  a'  representing  areas  of 
ends,  and  m  area  of  middle  section. 

Or,  (bxa-{-4mxn-\-dxc)xh^-6=zS,  a  b,  c  d  representing 
dimensions  of  ends,  and  m  n  of  the  middle  section. 

Note. — The  length  and  breadth  of  the  middle  section  are  respective- 
ly equal  to  half  the  sum  of  the  lengths  and  breadths  of  the  two  ends. 

Fig.  76.        a  b 


g  h 

Example. — What  are  the  contents  of  a  rectangular  pris- 
moid, Fig.  76,  the  lengths  and  breadths  of  the  two  ends  being 
7  and  6,  and  3  and  2  inches,  and  the  height  15  feet? 

7x6+3x2=42+6=48=sttm  of  the  areas  of  the  two  ends. 
7+3-i-2=10-7-2 =5 =length  of  the  middle  section. 
6+2-f-2=8-7-2=4  =  breadth  of  the  middle  section. 
5x4  X  4  =  80  =four  times  the"  area  of  the  middle  section. 


Then,  48  +  80X 


15ft. 
~6~ 


128  X  30=3840  cubic  inches. 


*  This  is  a  general  rule,  and  applies  equally  to  figures  of  proportion- 
ate or  dissimilar  ends.         v 


MENSURATION    OF    SOLIDS.  159 

Ex.  2.  What  is  the  capacity  of  a  prismoid,  the  ends  of  which 
are  respectively  6  by  8  and  9  by  12  inches,  and  the  height  of 
it  is  5  feet?*  Ana.  2.6389  feet. 

Ex.  3.  What  are  the  contents  of  a  prismoid  when  the  ends 
of  it  are  respectively  40.75  by  27.5  inches  and  20.5  by  14.75 
inches,  and  the  length  of  it  is  23.625  inches  ? 

Ans.  9.1392  cubic  feet 

Centre  of  Gravity.  For  rule,  see  Mensuration  of  Areas, 
Lines,  and  Surfaces,  p.  101. 

Wedge. 
Definition.     A  prolate  triangular  prism'. 

To  ascertain  the  Contents  of  a  Wedge,  Fig.  77. 
Rule. — To  the  length  of  the  edge,  e  g,  add  twice  the  length 
of  the  back ;  multiply  this  sum  by  the  perpendicular  height, 
ef,  and  then  by  the  breadth  of  the  back,  and  i  of  the  product 
will  be  the  solidity  required. 

Or,  (J+77x2xAx&)-f-6=S.       *   ' 
Fig.  77. 


Example. — The  back  of  a  wedge,  a  I,  d  c,  is  20  by  2  inches, 
and  its  height,  ef  20  inches;  what  are  its  contents'? 


20+20  X  2  =  60=length  of  the  edge  added  to  twice  the  length  of  the  back. 
60  X  20  X  2 =2400 —above  sum  multiplied  by  the  height,  and  that  product 
by  the  breadth  of  the  bach. 

2400-f-G=400=£  of  above  product  =  the  contents  required. 

*  An  excavation  or  embankment  of  a  road,  when  terminated  by  par- 
allel cross  sections,  is  a  rectangular  prismoid. 


160  MENSURATION    OF    SOUDS- 

Ex.  2.  The  back  of  a  wedge  is  15  inches  by  3  broad,  the 
edge  of  it  15  inches  in  length,  and  the  height  30;  what  are 
its  contents  in  inches?  Ans.  675  inches. 

Ex.  3.  The  back  of  a  wedge  is  64  inches  by  9  broad,  the 
length  of  the  edge  is  42  inches,  and  the  height  is  2  feet  4 
inches ;  how  many  cubic  feet  are  contained  in  it  ? 

Ans.  4.1319  cubic  feet 

Ex.  4.  The  height  of  a  wedge  is  15  inches,  the  edge  7,  and 
the  base  9  by  3 J ;  what  are  its  contents'? 

Ans.  218.75  cubic  inches. 

Centre  of  Gravity.  For  rule,  see  Mensuration  of  Areas, 
Lines,  and  Surfaces,  p.  100. 

Note. — "When  a  wedge  is  a  true  prism,  as  represented  by  Fig.  77, 
the  contents  of  it  are  equal  to  the  area  of  an  end  multiplied  by  its 
length. 

regular  bodies  (Polyhedrons). 

Definition.  A  regular  body  is  a  solid  contained  under  a  cer- 
tain number  of  similar  and  equal  plane  faces*  all  of  which  are 
equal  regular  polygons. 

Note  1.  The  whole  number  of  regular  bodies  which  can  possibly  be 
formed  is  five. 

Note  2.  A  sphere  may  always  be  inscribed  within,  and  may  always 
be  circumscribed  about  a  regular  body  or  polyhedron,  which  will  have  a 
common  centre. 

1.  The  Tetrahedron,  or  Pyramid,  Fig.  78,  which  has  four 
triangular  faces. 

2.  The  Hexahedron,  or  Cube,  Fig.  79,  which  has  six  square 
faces. 

S.^The  Octahedron,  Fig.  80,  which  has  eight  triangular  faces. 

4.  The  Dodecahedron,  Fig.  81,  which  has  twelve  pentagonal 
faces. 

5.  The  Icosahedron,  Fig.  82,  ivhich  has  twenty  triangular  faces. 

*  The  angle  of  the  adjacent  faces  of  a  polygon  is  called  the  diedral 
angle. 


MENSURATION   OF   SOLIDS. 


161 


If  the  following  figures  are  made  of  pasteboard,  and  the  lines  be  so 
cut  that  the  parts  may  be  turned  up  and  secured  together,  they  will  rep- 
resent the  five  bodies. 


Fig.  78. 


g.  71). 

Fig.  81. 


Fig.  80. 


Fig.  82. 


Tetrahedron. 

To  ascertain  the  Contents  of  a  Tetrahedron,  Fig.  83. 
Rule. — Multiply  -^  of  the  cube  of  the  linear  side  by  the 
square  root  of  2  (1.414213),  and  the  product  will  be  the  con- 
tents required. 

I3 
Or,  —x  -v/2z=S,  I  representing  the  length  of  a  side. 

Fig.  83. 


Example. — The  linear  side  of  a  tetrahedron,  a  be,  Fig.  83, 
is  4  ;  what  are  its  contents  ? 

43  G4 

jnX  V2=—  X  lAU  =  7.5U3=result  required. 


162  ^  MENSURATION   OP    SOLIDS. 

Ex.  2.  Required  the  contents  of  a  tetrahedron,  the  side  of 
which  is  6.  Ans.  25.452. 

Centre  of  Gravity.  Is  in  the  common  centre  of  the  centres 
of  gravity  of  the  triangles  made  by  a  section  through  the  cen- 
tre of  each  side  of  the  figures. 

Hexahedron,* 

To  ascertain  the  Contents  of  a  Hexahedron  {Cube),  see  Fig.  72, 
and  Rule,  p.  155. 

Octahedron. 
To  ascertain  the  Contents  of  an  Octahedron,  Fig.  84. 

Rule.— Multiply  •£-  of  the  cube  of  the  linear  side  by  the 
square  root  of.  2  (1.414213),  and  the  product  will  be  the  con- 
tents required. 


Or,  Z/xV2=S. 


Fig.  84 


Example. — What  are  the  contents  of  the  octahedron  abed, 

Fig.  84,  the  linear  side  of  which  is  4 1  ' 

43  64 

—  X  V  2  =— -  x  1.414  =30.1649  -result  required. 

o  o 

Ex.  2.  Required  the  contents  of  an  octahedron,  the  side  of 
which  is  8?  Ans.  241.3226. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

*  A  hexahedron  and  a  cube  are  identical  figures,  being  solids  having 
the  same  number  of  similar  and  equal  plane  faces. 


MENSURATION    OF    SOLIDS.  . .  163 

Dodecahedron. 
To  ascertain  the  Contents  of  a  Dodecahedron,  Fig.  85. 
Rule. — To  21  times  the  square  root  of  5  add  47,  and  di- 
vide the  sum  by  40 ;  then,  the  square  root  of  the  quotient  be- 
ing multiplied  by  5  times  the  cube  of  the  linear  side,  will  give 
the  contents  required. 

A/5x  21  +  47 * 

Qr'V  40  X*3*g=S. 

d 


Example. — The  linear  side  of  the  dodecahedron  abed  e*is 
3  ;  what  are  its  contents  1 

/•/5X21+47     __     _         /  2.23606X21 +47     J^ 
y/ — X27X5=W — X  135=  206.901  =  re- 
sult required. 

Ex.  2.  The  linear  side  of  a  dodecahedron  is  1 ;  what  is  the 
capacity  of  it?  4ns.  7.6631. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

Icosahedron. 

To  ascertain  the  Contents  of  an  Icosahedron,  Fig.  86. 

Eule. — To  3  times  the  square  root  of  5  add  7,  and  divide 

the  sum  by  2 ;  then,  the  square  root  of  this  quotient  being 

multiplied  by  -§  of  the  cube  of  the  linear  side,  will  give  the 

contents  required. 

V5x3+7     Px5 


°rV" 


=S. 


164  MENSURATION    OP    SOLIDS. 

Fig.  86.  d 


Example. — The  linear  side  of  the  icosahedron  ah  c  d  e  f 

is  3  ;  what  are  its  contents  % 

/-/5X3  +  7    33X5         /2.23606X3  +  7     27x5      , 
y/ T2 X~6-=V  2~§ X-g—v' 6.85409X22.5 

=58.9056 ^result  required. 

Ex.  2.  Required  the  contents  of  an  icosahedron,  the  linear 
side  of  which  is  1  *  Ans.  2. 1817. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 


REGULAR   BODIES. 

To  ascertain  the  Contents  of  any  regular  Solid  Body. 
•  When  the  Linear  Edge  is  given. 

Rule. — Multiply  the  cube  of  the  linear  edge  by  the  mul- 
tiplier in  column  A  in  the  table  on  the  following  page,  and 
the  product  will  be  the  contents  required. 

Example. — What  is  the  capacity  of  a  hexahedron  having 
sides  of  3  inches  ? 

33  X  1=27 =tabular  volume  multiplied  by  cube  of  edge  =contents  required. 

When  the  radius  of  the  Circumscribing  Sphere  is  given. 
Rule. — Cube  the  radius  of  the  circumscribing  sphere,  and 
multiply  it  by  the  multiplier  opposite  to  the  figure  in  column  B. 

Example. — The  radius  of  the  circumscribing  sphere  of  a 
hexahedron  is  1.732  inches;  what  is  the  volume  of  it? 

1.7323  X  1.5S06=product  of  cube  of  radius  and  tabular  multiplier =8  = 
result  required. 


MENSURATION   OF   SOLIDS. 


165 


When  the  radius  of  the  Inscribed  Sphere  is  given. 
Rule. — Cube  the  radius,  and  multiply  it  by  the  multiplier 
opposite  to  the  figure  in  column  C. 

Example. — The  radius  of  the  inscribed  sphere  of  a  hexahe- 
dron is  1  inch ;  what  is  its  volume  ? 

13  x  8  —product  of  cube  of  radius  and  tabular  multiplier =8  =result  re- 
quired. 


No.  of 
sides. 


Figures. 


A, 
By  linear 


B. 

By  radius 
of  circum. 
sphere. 


C. 

By  radius  of 
inscribing 
sphere. 


Angle  between 
two  adjacent 
faces. 


4 

6 

8 

12 

20 


Tetrahedron 

Hexahedron 

Octahedron 

Dodecahedron 

Icosahedron 


0.11785 

1. 

0.47140 

7.66312 

2.18169 


.51320 
1.53960 
1.33333 
2.78516 
2.53615 


13.85641 


70°  31' 42' 
90 
109  28  18 
5.55029116  33  54 
5.05406  138  11  23 


8. 
6.92820 


Note. — For  further  rules  to  ascertain  the  elements  of  polyhedrons, 
see  Appendix,  p.  262. 

Centre  of  Gravity.     Is  in  their  geometrical  centre. 

Cylinder. 

Definition.     A  figure  formed  by  the  revolution  of  a  right-an- 
gled parallelogram  around  one  of  its  sides. 

To  ascertain  the  Contents  of  a  Cylinder,  Fig.  87. 
Rule. — Multiply  the  area  of  the  base  by  the  height,  and 
the  product  will  give  the  contents  required. 

Or,  axh=zS. 

Fig.  87. 


166 


MENSURATION   OF    SOLIDS. 


Example. — The  diameter  of  a  cylinder,  b  c,  is  3  feet,  and 
its  length,  a  £,  7  feet ;  what  are  its  contents  ? 

Area  of  3  feet  =  7.068.     Then,  7.068  X  7=49.476 =result  required. 

Ex.  2.  "What  are  the  contents  of  a  cylinder,  the  height  of 
which  is  5  feet,  and  the  diameter  2  feet  ? 

Ans.  15.708  feet. 

Ex.  3.  The  circumference  of  the  base  of  a  cylindrical  col- 
umn is  20.42  feet,  and  the  height  of  the  column  is  9.695  feet; 
what  is  its  volume'?  Ans.  320.515  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 


Cone. 

Definition.     A  figure  described  by  the  revolution  of  a  right- 
angled  triangle  about  one  of  its  legs. 

To  ascertain  the  Contents  of  a  Cone,  Fig.  88. 
Eule. — Multiply  the  area  of  the  base  by  the  perpendicular 
height,  and  one  third  of  the  product  will  be  the  contents  re- 
quired. 

axh 


Or, 


3 


:S. 


Fig.  88. 


Example. — The  diameter,  a  b,  of  the  base  of  a  cone  is  15 
inches,  and  the  perpendicular  height,  c  d,  32.5  inches ;  what 
are  the  contents  of  the  cone  ? 


MENSURATION   OF    SOLIDS.  167 

Area  of  15  inches =176.7146. 

Then,  176,7UQ6X32-5  =  1914.4082  cubic  inches. 

Ex.  2.  The  diameter  of  the  base  of  a  cone  is  20,  and  its 
height  24  inches ;  what  are  its  contents  in  cubic  inches  % 

Ans.  2513.28  inches. 
Ex.  3.  What  are  the  contents  of  a  cone  when  the  diameter  of 
its  base  is  1.5  feet,  and  its  height  15  feet  ?     Ans.  8.8358  feet. 

Ex.  4.  The  diameter  of  the  base  of  a  cone  is  12.732  feet, 
and  the  height  of  it  is  50  feet ;  what  is  its  volume? 

Ans.  2121.9386  feet. 

Centre  of  Gravity.  It  is  at  a  distance  from  the  base  J  of 
the  line  joining  the  vertex  and  centre  of  gravity  of  the  base. 

To  ascertain  the  Contents  of  the  Frustrum  of  a  Cone,  Fig.  89. 

Rule.— Add  together  the  squares  of  the  diameters  of  the 
greater  and  less  ends  and  the  product  of  the  two  diameters ; 
multiply  their  sum  by  .7854,  and  this  product  by  the  height ; 
then  divide  this  last  product  by  three,  and  the  quotient  will 
give  the  contents  required. 

Or,  add  together  the  squares  of  the  circumferences  of  the  great- 
er and  less  ends  and  the  product  of  the  two  circumferences  ;  mul- 
tiply their  sum  by  .07958,  and  this  product  by  the  height;  then, 
divide  this  last  product"  by  three,  and  the  quotient  will  give  the  con- 
tents required. 

r\       79  i  j/9  .  3 — T/     -7854  x  A     ~ 

Or,  d2+d2+dxd'x ^ =S. 

o 
Fig.  89. 


168  MENSURATION   OF    SOLIDS. 

Example.-*— What  are  the  contents  of  the  frustrum  of  a 
cone,  the  diameters  of  the  greater  and  less  ends,  bd,ac,  being 
respectively  5  and  3  feet,  ami  the  perpendicular  height,  e  o, 
9  feet? 

52  +  32+5x3=49=^e  sum  of  the  squares  of  the  diameters  and  the 
product  of  the  diameters. 

49  X  .7854 =38.4846 =the  above  sum  by  .7854. 

— '■ — =1 15.4538  —  the  last  product  X  the  height  and  divided  by 

three,  which  is  the  result  required. 

Ex.  2.  What  are  the  contents  of  the  frustrum  of  a  cone, 
the  diameters  of  the  ends  being  respectively  2  ahd  4  feet,  and 
the  height  9  feet?  Ans.  65.9736  feet. 

Ex.  3.  The  frustrum  of  a  cone  is  12  inches  in  height,  and 
has  diameters  of  7  and  9^  inches ;  what  are  the  contents  of 
it?  Ans.  646.3842  inches. 

Centre  of  Gravity.     It  is  at  a  distance  from  the  base  £  of 

(R-f-r)2-l-2R2 
the  smaller  end— £  height  x-m  ,   ' — =-}  R  and  r  radii  of 

(R-j-r)2— Rr  " 

the  greater  and  less  ends. 


Pyramid. 

Definition.     A  figure,  the  base  of  which  has  three  or  more 
sides,  and  the  sides  of  which  are  plane  triangles. 

Note. — The  volume  of  a  pyramid  is  equal  to  one  third  of  that  of  a 
prism  having  equal  bases  and  altitude. 

To  ascertain  the  Contents  of  a  Pyramid,  Fig.  90. 
Rule. — Multiply  the  area  of  the  base  by  the  perpendicular 
height,  and  one  third  of  the  product  will  be  the  contents  re- 
quired. 

Or,  — =& 


MENSURATION   OF    SOLIDS. 
Fig.  90. 


169 


Example. — What  are  the  contents  of  a  hexagonal  pyramid, 
a  b  c,  Fig.  90,  a  side,  a  b}  being  40  feet,  and  its  height,  c  e, 
60  feet? 

40ax  2.5981  (tabular  multiplier,  p.  60)=4156.96=area  of  base. 

A  1  R.Ct  CiCt  \s  f*f\ 

'— =83139.2  —one  third  of  the  area  of  the  base  X  the  height= 

the  contents  required. 

Ex.  2.  The  height  of  a  quadrangular  pyramid  is  67  feet, 
and  the  width  of  its  base  is  16.5  feet;  what  are  its  contents 
in  cubic  feet?  Ans.  6080.25. 

Ex.  3.  What  are  the  contents  of  a  pentagonal  pyramid,  its 
height  being  12  feet,  and  each  of  its  sides  2  feet? 

Am.  27.528  feet. 

Centre  of  Gravity.  It  is  at  a  distance  from  the  base  i  of 
the  line  joining  the  vertex  and  centre  of  gravity  of  the  base. 

To  ascertain  the  Contents  of  the  Frustrum  of  a  Pyramid,  Fig.  91. 

Rule. — Add  together  the  squares  of  the  sides  of  the  great- 
er and  less  ends  and  the  product  of  these  two  sides ;  multiply 
the  sum  by  the  tabular  multiplier  for  areas  in  Table,  p.  60, 
and  this  product  by  the  height ;  then,  divide  the  last  prod- 
uct by  three,  and  the  quotient  will  give  the  contents  re- 
quired. 


Or,  (ss+s'z+s  x/)  X  tab.  mult.  x^=S}  s  and  sr  representing 

o 


the  lengths  of  the  sides. 


If 


170 


MENSURATION    OF    SOLIDS. 


Note. — When  the  areas  of  the  ends  are  known,  or  can  be  obtained 
without  reference  to  a  tabular  multiplier,  use  the  following. 


a-f  a/4-yaxa/XQ=S,  a  and  a' representing  areas  of  the 
o 


ends. 


Fig.  91. 


Example. — What  are  the  contents  of  the  frustrum  of  a  hex- 
agonal pyramid,  Fig.  91,  the  lengths  of  the  sides  of  the  great- 
er and  less  ends,  a  b,  c  d,  being  respectively  3.75  and  2.5  feet, 
and  its  perpendicular  height,  e  o,  7.5  feet  % 

3.752+2.52=20.3125=sww  of  the  squares  of  sides  of  greater  and  less 
ends. 

20.3125  +3.75  X  2.5 =29.6875  =above  sum  added  to  the  product  of  the 
two  sides. 

29.6875  X  2.5981  X  7.5  =578.48 —the  last  sum  X  tab.  mult.,  and  again  by 
the  height,  which,  -+3  =  192.82/eef. 

Ex.  2.  The  frustrum  of  a  hexagonal  pyramid  has  sides  of 
4  and  3  feet,  and  a  height  of  9  feet ;  what  are  its  contents  ! 

Ans.  294.3891  feet. 

When  the  Ends  of  a  Pyramid  are  not  those  of  a  Regular  Polygon, 
or  when  the  Areas  of  the  Ends  are  given. 
Rule. — Add  together  the  areas  of  the  two  ends  and  the 
square  root  of  their  product ;  multiply  the  sum  by  the  height, 
and  one  third  of  the  product  will  be  the  contents. 

Or,  a+a/+"V/axa/Xg=S. 


MENSURATION   OF    SOLIDS.  171 

*  Example. — What  are  the  contents  of  an  irregular-sided 
frustrum  of  a  pyramid,  the  areas  of  the  two  ends  being  22 
and  88  inches,  and  the  length  20  inches  ? 

22  +  88  =  110= sum  of  areas  of  ends. 

22x88  =  1936,  and  V 1936= 44  = square  root  of  product  of  areas. 

— =1026.66=one  third  of  sum  of  above  sum  and  product  X 

o 

the  height  —feet. 

Ex.  2.  The  areas  of  the  ends  of  an  irregular-sided  frustrum 
of  a  pyramid  are  81  and  100  inches,  and  the  length  25  inches; 
what  are  its  contents'?  Arts.  2258.33  inches. 

Centre  of  Gravity.     It  is  at  a  distance  from  the  centre  of 

the  smaller  end=i  heiqhtx\n — r^ — oc— 5  R  <*nd  r  radii  °f 

(R-f-r)2— Rr 

the  greater  and  less  ends. 

/Spherical  Pyramid. 
A  Spherical  Pyramid  is  that  part  of  a  sphere  included  within 
three  or  more  adjoining  plane  surfaces  meeting  at  the  centre  of  the 
sphere.  The  spherical  polygon  defined  by  these  plane  surfaces  of 
the  pyramid  is  called  the  base,  and  the  lateral  faces  are  sectors  of 
circles. 

To  ascertain  the  Elements  of  Spherical  Pyramids,  see  Docharty 
and  Hackley's  Geometry. 

Cylindrical  Ungulas. 

Definition.     Cylindrical  ungulas  are  frustrums  of  cylinders. 
Conical  ungulas  are  frustrums  of  cones* 

To  ascertain  the  Contents  of  a  Cylindrical  Ungula,  Fig.  92. 
1 .  When  the  section  is  parallel  to  the  axis  of  the  cylinder. 
Rule — Multiply  the  area  of  the  base  by  the  height  of  the 
cylinder,  and  the  product  will  be  the  contents  required. 
Or,  axh=S. 

*  For  Mensuration  of  Conical  Ungulas,  see  Conic  Sections,  p.  253. 


172 


MENSURATION    OF    SOLIDS. 

Fig.  92. 


Example. — The  area  of  the  base,  def  Fig.  92,  of  a  cylindric- 
al ungula  is  15.5  ins.,  and  its  height  20 ;  what  are  its  contents? 

15.5  X  20 =310  =product  of  area  and  height =result  required. 

Ex.  2.  The  area  of  the  base  of  a  cylindrical  ungula  is 
168.25  inches,  and  the  height  of  it  22 ;  what  are  its  contents 
in  cubic  feet?  Ans.  2.148  cubic  feet. 

2.  When  the  section  passes  obliquely  through  the  opposite  sides 
of  the  cylinder,  Fig.  93. 

Rule. — Multiply  the  area  of  the  base  of  the  cylinder  by 
half  the  sum  of  the  greatest  and  least  lengths  of  the  ungula, 
and  the  product  will  be  the  contents  required. 

Or,  axl+T'+2=S. 

Fig.  93. 


MENSURATION   OP    SOLIDS. 


173 


Example. — The  area  of  the  base  d  efof  a  cylindrical  un- 
gula  is  25  inches,  and  the  greater  and  less  heights  of  it,  a  d, 
b  e,  are  15  and  17  inches;  what  are  its  contents? 

25  X- —4:00=product  of half the  sum  of  the  heights  and  the  area 

of  the  base=result  required. 

Ex.  2.  The  area  of  the  base  of  a  cylindrical  ungula  is  75.8 
inches,  and  the  greater  and  less  heights  of  it  are  4.25  and  5.65 
feet ;  what  are  its  contents  in  cubic  feet  % 

Ans.  2.6056  cubic  feet. 

3.  When  the  section  passes  though  the  base  of  the  cylinder  and 
one  of  its  sides,  and  the  versed  sine  does  not  exceed  the  sine,  Fig.  94. 

Eule. — From  two  thirds  of  the  cube  of  the  sine,  a  d,  of  the 
arc,  dcfof  the  base,  subtract  the  product  of  the  area  of  the 
base  and  the  cosine,*  a  e,  of  the  half  arc. 

Multiply  the  difference  thus  found  by  the  quotient  arising 

from  the  height  divided  by  the  versed  sine,  and  the  product 

will  give  the  contents  required. 

s32     h 

Or,  -5 a xcX  — =S,  v  s  representing  the  versed  sine. 

o  v  s 

Fiff.di.    , 


Example. — The  sine  a  d  of  half  the  arc  of  the  base  of  an 
ungula,  Fig.  94,  is  5,  the  diameter  of  the  cylinder  is  10,  and 
the  height  of  the  ungula  10  ;  what  are  the  contents  of  it  ? 
*  When  the  cosine  is  0,  the  product  is  0. 


174 


MENSURATION    OF    SOLIDS. 


§•  of  53  =  83.333  =  taw  thirds  of  the  cube  of  the  sine. 

As  the  versed  sine  and  radius  of  the  base  are  equal,  the  cosine  is  0. 

Hence,  area  of  base  X  cosine =0. 

83.333  -OX-^ =166.666  ^difference  of  $  of  cube  of  the  sine  and  the 
product  of  area  of  base  and  the  cosine,  Xthe  height-— the  versed  sine=the 
contents  required. 

Ex.  2.  The  sine  of  half  the  arc  of  the  base  of  an  ungula  is 
12  inches,  the  diameter  of  the  cylinder  is  25,  and  the  height 
of  the  ungula  18 ;  what  are  its  contents  ? 

Ans.  1190.34375  inches. 
For  rules  to  ascertain  the  area  of  base,  see  pp.  85-87. 

4.  When  the  section  passes  through  the  base  of  the  cylinder, 
and  the  versed  sine  exceeds  the  sine,  Fig,  95. 

Rule. — To  two  thirds  of  the  cube  of  the  sine  of  half  the 
arc  of  the  base,  add  the  product  of  the  area  of  the  base  and 
the  excess  of  the  versed  sine  over  the  sine  of  the  base. 

Multiply  the  sum  thus  found  by  the  quotient  arising  from 

the  height  divided  by  the  versed  sine,  and  the  product  will  be 

the  contents  required. 

^     2  s3,  h 

Or,  -j-+ax(vscvs)X  — =S. 


Fig.  95. 


Example. — The  sine  a  d  of  half  the  arc  of  an  ungula,  Fig.  95, 
is  12  inches,  the  versed  sine  a  g  is  16,  the  height  c  g  20,  and 
the  diameter  of  the  cylinder  25  inches ;  what  are  the  contents? 


MENSURATION   OF   SOLIDS. 


175 


§■  of  123  =  ll52  =  two  thirds  of  cube  of  sine  of  the  arc  of  the  base. 

Area  of  base  (see  Rules,  p.  84-134) =331.78. 

1152+(331.78xl6-12.5)=2313.23=swm  of  %  of  the  cube  of  the  sine 
of  the  base,  and  product  of  area  of  base,  and  difference  between  the  versed 
sine  and  sine  of  the  base. 

2313.23  X  20-T-16  =  2891.5375  =product  of  above  sum  and  the  height,  di- 
vided by  the  versed  sine=result  required. 

5.  When  the  section  passes  obliquely  through  both  ends  of  the 
cylinder,  Fig.  96. 

Rule. — Conceive  the  section  to  be  continued  till  it  meets 
the  side  of  the  cylinder  produced ;  then,  as  the  difference  of 
the  versed  sines  of  the  arcs  of  the  two  ends  of  the  ungula  is  to 
the  versed  sine  of  the  arc  of  the  less  end,  so  is  the  height  of  the 
cylinder  to  the  part  of  the  side  produced. 

Find  the  contents  of  each  of  the  ungulas  by  rules  3  and  4, 
and  their  difference  will  be  the  contents  required. 

v'  X  h 

Or,  — — -p  =  h\  v  and  v'  representing  the  versed  sines  of  the 

arcs  of  the  two  ends,  h  the  height  of  the  cylinder,  and  li  the  height 
of  the  part  produced. 

Fig.  96.  k/ 


Example. — The  versed  sines,  a  e,do,  and  sines,  %  h,  g  r,  of 
the  arcs  of  the  two  ends  of  an  ungula,  Fig.  96,  are  assumed  to 
be  respectively  8.5  and  25,  and  11.5  and  0  inches,  the  length 
of  the  ungula  within  the  cylinder,  cut  from  one  having  25 
inches  diameter  is  20  inches ;  what  is  the  height  of  the  un- 


176  MENSURATION   OF    SOLIDS. 

gula  produced  beyond  the  cylinder,  and  what  the  contents  of 
the  ungula? 

25<X)8.5  :  8.5 : :  20 :  10.303 =height  of  ungula produced  beyond  the  cylinder. 

Lower  ungula,  the  sine,  g  r,  being  0,  the  versed  sine = the  diameter. 

Base  of  ungula  being  a  circle  of  25  inches  diameter,  area=490.874. 

The  versed  sine  and  diameter  of  the  base  being  equal  (25),  the  sine  is  0. 

490.874  X  25  =6135.925  =/>rodWtf  of  area  of  base  and  excess  of  versed 
sine  over  the  sine  of  the  base. 

30.303-4-25  =  1.2121  —quotient  of  heights-versed  sine. 

Then,  6135.925X1. 2121 =7  437. 35  i7=product  of  above  product  and 
quotient=the  residt  required. 


Definition.     A  solid,  the  surface  of  which  is  at  a  uniform  dis- 
tance from  the  centre. 

To  ascertain  the  Contents  of  a  Sphere,  Fig.  97. 
Rule. — Multiply  the  cube  of  the  diameter  by  .5236,  and 
the  product  will  be  the  contents  required. 

Or,  d3x.5236=S,  d  representing  the  diameter. 

7-7.7.  97.  c 


d 
Example. — What  are  the  contents  of  a  sphere,  Fig.  97,  its 
diameter,  a  b,  being  10  inches? 

103=1000>  and  1000  x  .5236=523.6  cubic  inches. 

Ex.  2.  The  diameter  of  a  sphere  is  17  inches ;  what  are  its 
contents'?  Ans.  1.4887  cubic  feet. 

Ex.  3.  What  are  the  contents  of  a  globe  10.5  feet  in  di- 
ameter? Ans.  606.132  cubic  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

Note.— .5236=f  of  3.1416. 


MENSURATION   OF    SOLIDS.  177 

Segment  of  a  Sphere. 
Definition*     A  section  of  a  sphere. 

To  ascertain  the  Contents  of  a  Segment  of  a  Sphere,  Fig.  98. 
Rule  1.  To  three  times  the  square  of  the  radius  of  its  base 
add  the  square  of  its  height ;  multiply  this  sum  by  the  height, 
and  the  product,  multiplied  by  .5236,  will  give  the  contents 
required. 

Or,  (3r2-f/*2)x^X.5236=S. 

2.  From  three  times  the  diameter  of  the  sphere  subtract 
twice  the  height  of  the  segment ;  multiply  this  remainder  by 
the  square  of  the  height,  and  the  product,  multiplied  by  .5236, 
will  give  the  contents  required. 

Or,  3  d-2  hxh2x.523Q=S. 

0 

Fig.  98. 


Example. — The  segment  of  a  sphere,  Fig.  98,  has  a  radius, 
a  o,  of  7  inches  for  its  base,  and  a  height,  b  o,  of  4  inches ; 
what  are  its  contents ? 

72  X  3 +4?  =  163= the  sum  of  three  times  the  square  of  the  radius  and  the 
square  of  the  height. 

1G3  X  4  X  .5236 =341 .3872 =the  above  sum  multiplied  by  the  height,  and 
by  .5236=inches. 

Ex.  2.  The  radius  of  a  spherical  segment  is  48  inches, 
and  the  height  12  inches;  what  are  its  contents? 

Ans.  44334.2592  cubic  inches. 
Ex.  3.  The  height  of  a  spherical  segment  is  2  inches,  and 
the  diameter  of  the  sphere  6  inches ;  what  are  its  contents? 

Ans.  29.322  cubic  inches. 
112 


178  MENSURATION   OF    SOLIDS. 

Centre  of  Gravity.    Distance  from  the  centre,  3.1416  v2lr— -J 

—s,  v  being  the  versed  sine,  s  the  contents  of  the  segment,  and  r 
the  radius  of  the  sphere. 

=•5-* — TTjh,  h  representing  the  height 
or  versed  sine  of  the  segment. 

Of  a  Hemisphere.     Distance  from  the  centre  -§  r. 

Spherical  Zone  {or  Frustrum  of  a  Sphere). 

Definition.     The  part  of  a  sphere  included  between  two  paral- 
lel chords. 

To  ascertain  the  Contents  of  a  Spherical  Zone,  Fig.  99. 

Rule. — To  the  sum  of  the  squares  of  the  radii  d  c  and/ g 

of  the  two  ends,  df  eh,  add  one  third  of  the  square  of  the 

height,  c  g,  of  the  zone ;  multiply  this  sum  by  the  height,  and 

again  by  1.5708,  and  it  will  give  the  contents  required. 

h2 
Or,  r2+r/2+-XAxl.5708=S. 


Fig.  99. 


/ 

Example. — What  are  the  contents  of  a  spherical  zone,  Fig. 
99,  the  greater  and  less  diameters,  /  h  and  d  e,  being  20  and 
15  inches,  and  the  distance  between  them,  or  height  of  the 
zone,  being  10  inches. 

102  +  7.52  =  1 56.25  =sum  of  the  squares  of  the  radii  of  the  two  ends. 

102 
156.25+—=:  189.58 —the  above  sum  added  to  one  third  of  the  square 
o 
of  the  height. 

189.58  X 10  X  1.5708 =2977.9226 —the,  last  sum  multiplied  by  the  height 
and  again  by  1.5708  =feet. 


MENSURATION    OF    SOLIDS. 


179 


Ex.  2.  A  zone  of  a  sphere  has  the  radii  of  its  ends  each  6 
inches,  and  its  height  is  8  inches ;  what  are  its  contents  ? 

Ans.  1172.86  cubic  inches. 

Ex.  3.  What  are  the  contents  of  the  zone  of  a  sphere,  the 
radii  of  its  ends  being  10  and  12  inches,  and  the  height  of  it 
4  inches?  Ans.  1566.6112  cubic  inches. 

Centre  of  Gravity.    Right  Zone.    Is  in  its  geometrical  centre. 

Of  a  Frustrum.    — —  =d,  representing  the  distance  from  the 

lor — 4A 

vertex  of  the  frustrum. 

Spheroids  (Ellipsoids). 

Definition.  Solids  generated  by  the  revolution  of  a  semi-ellipse 
about  one  of  its  diameters. 

When  the  revolution  is  about  the  transverse  diameter,  they  are 
Prolate,  and  when  about  the  conjugate,  they  are  Oblate. 

To  ascertain  the  Contents  of  a  Spheroid,  Figs.  100  and  101. 

Rule. — Multiply  the  square  of  the  revolving  axis,  c  d,  by 
the  fixed  axis,  a  b,  and  this  product  by  .5236,  and  it  will 
give  the  contents  required. 

Or  a2  x  at  X  -5236  =  S,  a  and  a'  representing  the  axes. 

4 
Or,  -  3.1±16  xr2  xr'=S,  r  and  ?%/  representing  the  semi-axes. 

o 

Note. — The  contents  of  a  spheroid  are  equal  to  two  thirds  of  a  cylin- 
der that  will  circumscribe  it. 


Fig.  100. 


Fig.  101. 


it? 


Example. — In  a  prolate  spheroid,  Fig.  100,  the  fixed  axis 
a  b  is  14,  and  the  revolving  axis  c  d  10 ;  what  are  its  con- 
tents? 

102X  14  =  1400  ^product  of  square  of  revolving  axis  and  fixed  axis. 

1400  X  .5236 =733.04 = above  product  by  .523S=result  required. 


180  MENSURATION    OF    SOLIDS. 

Ex.  2.  The  axes  of  a  prolate  spheroid  are  100  and  60 
inches  ;  what  are  its  contents'?  Ans.  188496  inches. 

Ex.  3.  The  axes  of  an  oblate  spheroid,  Fig.  101,  are  10  and 
14  inches ;  what  are  its  contents'?       Ans.  1026.256  inches. 

Ex*  4.  What  are  the  contents  of  an  oblate  spheroid,  its 
transverse  axis,  c  d,  being  24,  and  its  conjugate,  a  b,  18  inches  ! 

Ans.  5428.685  inches. 

Ex.  5.  What  are  the  contents  of  an  oblate  spheroid,  the  axes 
of  which  are  50  and  30  inches  t       Ans.  22.7257  cubic  feet. 

Centre  of  Gravity.     Is  in  their  geometrical  centre. 

Segments  of  Spheroids. 

To  ascertain  the  Contents  of  the  Segment  of  a  Spheroid. 

When  the  base,  e  f,  is  Circular,  or  parallel  to  the  revolving 
cms,  as  c  d,  Figs.  102  and  102*. 

Rule. — Multiply  the  fixed  axis,  a  b,  by  three,  the  height  of 
the  segment,  a  o,  by  two,  and  subtract  the  one  product  from 
the  other ;  multiply  the  remainder  by  the  square  of  the  height 
of  the  segment,  and  the  product  by  .5236. 

Then,  as  the  square  of  the  fixed  axis  is  to  the  square  of  the 
revolving  axis,  so  is  the  last  product  to  the  contents  of  the 
segment. 

Or, j— '• =S,  a  and  a/  representing 

the  fixed  and  revolving  axes. 
Fig.  102.  c  Fig.  102*.  a 


Example. — In  a  prolate  spheroid,  Fig.  102,  the  fixed  or 
transverse  axis,  a  b,  is  100,  and  the  revolving  or  conjugate, 
c  d,  GO;  what  are  the  contents  of  a  segment  of  it,  its  height, 
a  o,  being  10  inches  ? 


MENSURATION   OF   SOLIDS. 


181 


100x3  —  10X2  =280=twice  the  height  of  the  segment  subtracted  from 
three  times  the  fixed  axis. 

280  X  102X. 5236  =  14660. 8  =product  of  above  remainder,  the  square  of 
the  height,  and  .5236. 

Then,  1002:  602::  H660.8:  5277.888 =the  result  required. 

Ex.  2.  The  height  of  a  segment  of  a  prolate  spheroid,  Fig. 
102,  is  5  inches ;  what  are  its  contents,  the  transverse  axis 
being  4  feet  2  inches,  and  the  conjugate  2.5  feet? 

Ans.  659.736  cubic  inches. 
Ex.  3.  The  height  of  a  segment  of  an  oblate  spheroid,  Fig. 
102*,  is  10  inches,  the  transverse  diameter  being  100,  and 
the  conjugate  60;  what  are  its  contents? 

Ans.  23271.111  cubic  inches. 


When  the  base,  e  ft  is  Elliptical,  or  perpendicular  to  the  revolv- 
ing axis,  as  c  d,  Figs.  103  and  103*. 

Eule. — Multiply  the  revolving  axis,  c  d,  by  three,  the  height 
of  the  segment,  c  o,  by  two,  and  subtract  the  one  from  the 
other ;  multiply  the  remainder  by  the  square  of  the  height 
of  the  segment,  and  the  product  by  .5236. 

Then,  as  the  revolving  axis  is  to  the  fixed  axis,  so  is  the 
last  product  to  the  contents. 

Or,  * ~— - ' =  S ;  a  representing  the  fixed 

a 

and  af  the  revolving  axes. 
Fig.  103.  c  Fig.  103*. 


Example. — The  diameters,  c  d  and  a  b,  of  an  oblate  sphe- 
roid, Fig.  103*,  are  100  and  60  inches,  and  the  height  of  a 
segment,  c  o,  thereof  is  12  inches ;  what  are  its  contents  ? 


182 


MENSURATION    OF    SOLIDS. 


100x3— 12x2  =276=  twice  the  height  of  the  segment  subtracted  from 
three  times  the  revolving  axis. 

276  X  122  X  .5236  -20809.9584  -product  of  above  remainder,  the  square 
of  the  height,  and  .5236. 

Then,  100:  60:: 20809.9584 :  12485.975  =  ^6  result  required. 

Ex.  2.  The  segment  of  a  prolate  spheroid,  Fig.  103,  is  20 
inches  in  height,  the  revolving  diameter  of  the  spheroid  being 
10  feet,  and  the  fixed  axis  16  feet  8  inches;  what  are  its  con- 
tents in  cubic  inches.  Ans.  111701.333  inches. 

Ex.  3.  The  segment  of  an  oblate  spheroid,  Fig.  103*  is  2 
feet,  and  the  axes  of  the  spheroid  are  200  and  120  inches ; 
what  are  its  contents  in  cubic  feet  ?  Ans.  57 .8054  feet. 

Frustra  of  Spheroids. 

To  ascertain  the  Contents  of  the  Middle  Frustrum  of  a  Spheroid. 

When  the  ends,  e  f  and  g  h,  are  circular,  or  parallel  to  the 
revolving  axis,  as  c  d,  Figs.  104  and  104*. 

Rule. — To  twice  the  square  of  the  revolving  axis,  c  d,  add 
the  square  of  the  diameter  of  either  end,  eforg  h;  multiply 
this  sum  by  the  length,  {  o,  of  the  frustrum,  and  the  product 
again  by  .2618,  and  it  will  give  the  contents  required. 

Or,  2a/2+d2x/x.2618=S. 


Fig.  104. 


Fig.  104 


Example. — The  middle  frustrum,  i  o,  of  a  prolate  sphe- 
roid, Fig.  104,  is  36  inches  in  length,  the  diameters  of  it 
being,  in  the  middle,  c  d,  50  inches,  and  at  its  ends,  e  f  and 
g  h,  40 ;  what  are  its  contents  ? 

502X  2 +402 =6600 =sum  of  twice  the  square  of  the  middle  diameter 
added  to  the  square  of  the  diameter  of  the  ends. 

6600x36x.2618=62203.68=^ro(fMc/  of  the  above  sum,  the  length  of 
the  frustrum,  and  .2618  =  ^e  result  required. 


MENSURATION    OF    SOLIDS. 


183 


Ex.  2.  What  are  the  contents  of  the  middle  frustrum  of  an 
oblate  spheroid,  Fig.  104*,  the  transverse  diameter  being  100 
inches,  the  diameters  of  the  ends  of  the  frustrum  each.  80 
inches,  and  the  length  of  it  3  feet  4  inches'? 

Ans.  276460.8  cubic  inches. 

Ex.  3.  The  middle  frustrum  of  a  prolate  spheroid,  Fig.  104, 
is  80  inches  in  length,  the  diameter  of  it  being,  in  the  middle, 
60  inches,  and  at  its  ends  38.5  inches;  what  are  its  contents 
in  cubic  feet?  Ans.  105.232  cubic  feet. 

Ex.  4.  The  diameter  of  the  middle  frustrum  of  a  prolate 
spheroid,  Fig.  104,  is  100  inches,  that  of  the  end  of  the  frus- 
trum is  60  inches,  and  the  length  of  it  is  8  feet ;  what  are 
its  contents'?  Ans.  593134.08  cubic  inches. 

When  the  ends,  e  f  and  g  h,  are  elliptical,  or  perpendicular  to 
the  revolving  axis,  as  c  d,  Figs.  1 05  and  1 05*. 

Eule.— To  twice  the  product  of  the  transverse  and  conju- 
gate diameters  of  the  middle  section,  a  b,  add  the  product  of 
the  transverse  and  conjugate  of  either  end,  ef  or  g  h;  multi- 
ply this  sum  by  the  length,  o  i,  of  the  frustrum,  and  the  prod- 
uct by  .2618,  and  it  will  give  the  contents  required. 

Or,  a/xax2-fO<?x/x.2618=S. 
Fig.  105.  c  Fig.  105*.  a 


d  b 

Example. — In  the  middle  frustrum  of  a  prolate  spheroid, 
Fig.  105,  the  diameters  of  its  middle  section  are  50  and  30 
inches,  and  of  its  ends  40  and  24  inches ;  what  are  its  con- 
tents, its  length  being  18  inches? 

50  X  30  X  2 =3000 =twice  the  product  of  the  transverse  and  conjugate  di- 
ameters. 


3000+40  X  24  = 3960 =sum  of  the  above  product  and  the  product  of  the 
transverse  and  conjugate  diameters  of  the  ends. 

3960 X 18 X. 2618  =  18661. 104=^e  preceding  product  into  the  length 
.2618  —  the  contents  required. 


184  MENSURATION   OF    SOLIDS. 

Ex.  2.  The  diameters  of  the  middle  frustrum  of  an  oblate 
spheroid,  Fig.  105*  are  100  and  60  inches  in  the  middle,  and 
60  and  40  inches  at  the  ends,  and  the  length  of  it  is  6  feet  8 
inches ;  what  are  its  contents  ?     Arts.  301593.6  cubic  inches. 

Ex.  3.  The  middle  frustrum  of  a  prolate  spheroid  is  3  feet 

4  inches  in  height,  and  has  diameters  of  8  feet  4  inches  and 

5  feet  in  its  middle,  and  of  6  feet  8  inches  and  4  feet  at  its 
ends ;  what  are  its  contents  in  cubic  feet  ? 

Ans.  95.993  cubic  feet. 
Centres  of  Gravity.     Frustrum.     See  Appendix,  p.  282. 
Middle  Frustra,  or  Zones.     Is  in  their  geometrical  centre. 
Semi-spheroids. — Distance  from  the  centre  of  the  spheroid. 
Prolate,  fa.      Oblate.  f  5. 

(a-4-d)2  (b  +  d')2 

Segments. — Prolate.  §  — -.  Oblate,  f  — — — r,  b  repre- 
senting the  semi-conjugate  axis,  a  the  semi-transverse,  and  d,  d/  the 
distances  of  the  base  of  the  segments  from  the  centre  of  the  spheroid. 

Cylindrical  Ring. 
Definition.     A  ring  formed  by  the  curvature  of  a  cylinder. 

To  ascertain  the  Contents  of  a  Cylindrical  Ring,  Fig.  106. 

Rule. — To  the  diameter  of  the  body  of  the  ring,  a  b,  add 
the  inner  diameter  of  the  ring,  be;  multiply  the  sum  by  the 
square  of  the  diameter  of  the  body,  the  product  by  2.4674,  and 
it  will  give  the  contents  required. 

Or,  d-\-d'  xd7  x  2.4674  =rS,  d  and  d/  representing  the  diam- 
eter of  the  body  and  inner  diameter. 

Or,  a  x  £=S,  a  representing  area  of  section  of  body,  and  I  the 
length  of  the  axis  of  the  body. 

Fig.  106.        ^~ZT^\ 


MENSURATION   OF    SOLIDS.  185 

Example. — What  are  the  contents  of  an  anchor-ring,  the 
diameter  of  the  metal  being  3  inches,  and  the  inner  diameter 
of  the  ring  8  inches  % 

,  3  +  8x3*=9d=p?'oduct  of  sum  of  diameters  and  the  square  of  diameter 
of  body  of  ring. 

99  x2A674:=24:4:.2726=above product  X 2.4674  =  *//e  contents  required. 

Ex.  2.  The  diameter  of  the  body  of  a  cylindrical  ring  is  2 
inches,  and  the  inner  diameter  of  the  ring  is  12  inches;  what 
are  its  contents'?  Ans.  138.1744  cubic  inches. 

Ex.  3.  The  dimensions  of  a  cylindrical  ring  are  14  inches 
in  diameter  of  body  and  16  inches  in  the  ring ;  what  are  its 
contents'?  Ans.  14508.312  cubic  inches. 

Centre  of  Gravity.     Is  in  its  geometrical  centre.  . 


LINKS. 

Definition.     Elongated  or  elliptical  rings. 

Elongated  or  Elliptical  Links. 

To  ascertain  the  Contents  of  an  Elongated  or  Elliptical  Link, 
Figs.  107  and  108. 

Rule. — Multiply  the  area  of  a  section  of  the  body,  a  b, 
of  the  link  by  its  length,  or  the  circumference  of  its  axis,  and 
the  product  will  give  the  contents  required. 

Or,  a  x  1=  S,  a  representing  area  of  section  of  body,  and  I  the 
length  of  the  axis  of  the  body. 

Note. — By  rule,  p.  Ill,  the  circumference  or  length  of  the  axis 
of  an  elongated  link=the  sum,  of  3.1416  times  the  sum  of  the  less 
diameter  added  to  the  thickness  of  the  ring,  and  the  product  of  twice 
the  remainder  of  the  less  diameter  subtracted  from  the  greater. 

Note. — By  rule,  p.  Ill,  the  circumference  or  length  of  the  axis  of 
an  elliptical  ring  =  the  square  root,  of  half  the  sum  of  the  diameters 
squared  x  3.1 4 1G. 


18G 


MENSURATION   OF   SOLIDS. 

Fig.  107.  Fig.  108 


Example. — The  elongated  link  of  a  chain,  Fig.  107,  is  1 
inch  in  diameter  of  body,  and  its  inner  diameters,  b  c  and  e  f, 
are  10  and  2.5  inches ;  what  are  its  contents? 

Area  of  1  inch  =  .7854. 

2.5  +  1x3.1416  =  10.9956=3.1416  times  the  swn  of  the  less  diameter 
and  thickness  of  the  ring  =  length  of  axis  of  ends. 

10— 2.5x2  =  15  =  twice  the  remainder,  of  the  less  diameter  subtracted 
from  the  greater =length  of  sides  of  body. 

Then,  10.9956  +  15  =25.9956  =  length  of  axis  of  link. 

Hence,  .7854x25.9956=20.417=area  of  link  X  its  length=result  re- 
quired. 

Ex.  2.  The  elongated  link  of  a  chain  cable  is  1.5  inches 
in  diameter  of  body,  and  its  inner  diameters  are  3.5  and  4.5 
inches;  what  are  its  contents?  Ans.  38.8584  inches. 

Ex.  3.  The  elliptical  link  of  a  chain,  Fig.  108,  is  1  inch  in 
diameter,  a  b,  of  body,  and  its  inner  diameters,  b  c  and  e  f, 
are  10  and  2.5  inches;  what  are  its  contents  ? 


v/ 


2.5  +  1  +10  +  1  =133. 25=  diameters  of  axes  squared. 
133.25 


X  3. 1416=25. 6iS=square  root  of  diameters  squared X  3.141 6 

= circumference  of  axis  of  ring. 
Area  of  1  inch =.7854. 
Then,  25.643x.7854=20.14=reswft  required. 

Ex.  4.  An  elliptical  link  has  diameters  of  11.5  and  5.5 
inches,  its  diameter  of  body  being  2.5  inches;  what  are  its 
contents?  Ans.  185.8272  inches. 

Centres  of  Gravity.     Is  in  their  geometrical  centres. 


MENSURATION   OP   SOLIDS.  187 

Spherical  Sector. 

Definition.  A  figure  generated  by  the  revolution  of  a  sector 
of  a  circle  about  a  straight  line  drawn  through  the  vertex  of  the 
sector  as  an  axis. 

Note. — The  arc  of  the  sector  generates  the  surface  of  a  zone  termed 
the  hase  of  the  sector  of  a  sphere,  and  the  radii  generates  the  surfaces 
of  two  cones,  having  a  vertex  in  common  with  the  sector  at  the  centre 
of  the  sphere. 

To  ascertain  the  Contents  of  a  Spherical  Sector,  Fig.  109. 
Rule. — Multiply  the  surface  of  the  zone,  which  is  the  base  of 
the  sector,  by  one  third  of  the  radius  of  the  sphere,  and  the 
product  will  give  the  contents  required. 

r 
Or,  ax-z=S,a  representing  the  area  of  the  base. 

Fig.  109. 


si ■    u^- 

Example. — What  are  the  contents  of  a  spherical  sector, 
Fig.  109,  generated  by  the  sector  c  a  h,  the  height  of  the  zone 
abed  being,  a  o,  12  inches,  and  the  radius,  g  h,  of  the  sphere 
15  inches? 

12  X  94.248  =  1130.976  =height  of  zone  X  circumference  of  sphere=sur- 
face  of  zone  (see  p.  90). 

1130.976  X  ^=5654.88  =product  of  surf  ace  of  zone  and  £  of diameter 
(—$  of  radius)  =  result  required. 

Ex.  2.  The  diameter  of  a  sphere  is  10  inches,  and  the  height 
of  the  zone  or  base  of  a  sector  is  5  inches ;  what  are  its  con- 
tents? Ans.  261.5  inches. 

Centre  of  Gravity.     Distance  from  the  centre =j(r— JA). 
Distance  from  the  vertex = — - — ,  h=zthe  height  of  the  zone. 

o 


188  MENSURATION    OF    SOLIDS. 

Note. — The  surface  of  a  spherical  sector=the  sum  of  the  areas  of 
the  zone  and  the  two  cones. 


SPINDLES. 

Definition.  Figures  generated  by  the  revolution  of  a  plane 
area  bounded  by  a  curve,  when  the  curve  is  revolved  about  a  chord 
perpendicular  to  its  axis,  or  about  its  double  ordinate,  and  they  are 
designated  by  the  name  of  the  arc  from  which  they  are  generated, 
as  Circular,  Elliptic,  Parabolic,  etc. 

Circular  Spindle. 

To  ascertain  the  Contents  of  a  Circular  Spindle,  Fig.  110. 

Rule. — Multiply  the  central  distance,  o  e,  by  half  the  area 
of  the  revolving  segment,  a  c  ef  Subtract  the  product  from 
one  third  of  the  cube  of  half  the  length,/  e;  then  multiply  the 
remainder  by  12.5664,  and  the  product  will  give  the  contents 
required. 

a     (1—2)3 
Or,  c  x  r ~-  X  12.5664  =  S,  I  representing  the  length  of 

the  spindle,  and  a  the  area  of  the  revolving  segment. 

Fig.  110.  a 


Example. — What  are  the  contents  of  a  circular  spindle, 
when  the  central  distance,  o  e,  Fig.  110,  is  7.071067  inches, 
the  length,/*?,  14.14213,  and  the  radius,  o  c,  10  inches? 

Note. — The  area  of  the  revolving  segment, /e,  being=the  side  of  the 
square  that  can  be  inscribed  in  a  circle  of  20,  is  202x  .7854  — 14.14213* 
-H4=28.54. 


MENSURATION    OF    SOLIDS.  189 

7.071067 Xl4.27=100.9041=cen*ra/  distanceX half  area  of  revolving 
segment. 

7 071673 

100.9041 '—— =  16.947= remainder  of  above  product  and  i  of 

o 
cube  of  half  the  length. 

16.947  X  12.5664=212.9628  =result  required. 

Ex.  2.  The  central  distance  of  a  circular  spindle  is  3  inches, 
the  area  of  the  revolving  segment  is  11.5  inches,  and  the  length 
of  the  spindle  8  inches ;  what  are  the  contents  of  it  ? 

Ans.  51.3086  inches. 
Ex.  3.  The  length  of  a  circular  spindle  is  24  inches,  and  its 
diameter  18  ;  what  are  the  contents  of  it? 

Ans.  3739.585  inches. 

The  chord  of  the  arc,  24,  and  the  versed  sine  C1^),  9,  being  given,  the 
diameter  of  the  circle  is  found  by  rules  p.  75,  76. 


v© 


2 

+92  =  15=  chord  of  half  the  arc. 


152 
Hence  -—■ = 25  =  diameter. 

Area  of  revolving  segments,  per  table,  p.  136,  137=159.094. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

Frustrum  or  Zone  of  a  Circular  Spindle.* 

To  ascertain  the  Contents  of  a  Frustrum  or  Zone  of  a  Circular 
Spindle,  Fig.  111. 
Rule. — From  the  square  of  half  the  length,  /  g,  of  the 
whole  spindle,  take  |  of  the  square  of  half  the  length,  t  c,  of  the 
frustrum,  and  multiply  the  remainder  by  the  said  half  length 
of  the  frustrum  ;  multiply  the  central  distance,  o  c,  by  the  re- 
volving area  which  generates  the  frustrum;  subtract  this 
product  from  the  former,  and  the  remainder,  multiplied  by 
6.2832,  will  give  the  contents  required. 

Note. — The  revolving  area  of  the  frustrum  can  be  obtained  by  di- 
viding its  plane  into  a  segment  of  a  circle  and  a  parallelogram. 

*  The  middle  frustrum  of  a  circular  spindle  is  one  of  the  various 
forms  of  casks. 


190  MENSURATION   OF    SOLIDS. 

2 

Or,  7^2-l-^x~(cxa)xG.2832  =  S,  I  and  f  repre- 

senting  the  lengt/is  of  the  spindle  and  of  the  frustrum,  a  area  of 
the  revolving  section  of  frustrum,  and  c  the  central  distance. 


Example. — The  length  of  the  middle  frustrum  of  a  circu- 
lar spindle,  I  c,  is  6  inches,,  the  length  of  the  spindle,  fg,i$  8 
inches,  the  central  distance,  o  e,  is  3  inches,  and  the  area  of  the 
revolving  or  generating  segment  is  10  inches ;  what  are  the 
contents  of  the  frustrum  1 

(8+2)*-^— J- =  13x3 =39  =product  of  half  the  length  of  the  frus- 
3 
trurn  and  the  remainder  of%  the  square  of  half  the  length  of  the  frustrum 
subtracted  from  the  square  of  half  the  length  of  the  spindle. 

39— 3  X  10= 9  =product  of  the  central  distance  and  the  area  of  the  seg- 
ment subtracted  from  preceding  product. 

9  X  6.2832 =56.5488 = last  product  X  6.2832 =result  required. 

Ex.  2.  The  length  of  a  circular  spindle  is  1.333  feet,  the 
length  of  the  middle  frustrum  of  it  is  1  foot,  the  central  dis- 
tance is  6  inches,  and  the  area  of  the  revolving  segment  is  40 
inches ;  what  are  the  contents  of  the  frustrum  *? 

Ans.  452.3904  cubic  inches. 

Ex.  3.  The  length  of  the  middle  frustrum  of  a  circular  spin- 
dle is  12  inches,  the  length  of  the  spindle  being  24,  the  central 
distance  is  3.5,  and  the  area  of  the  revolving  segment  is  96 
inches ;  what  are  the  contents  of  the  frustrum  ? 

Ans.  2865.1392  cubic  inches. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 


MENSURATION   OF    SOLIDS.  191 

Segment  of  a  Circular  Spindle. 

To  ascertain  the  Contents  of  a  Segment  of  a  Circular  Spindle, 
Fig.  112. 

Rule. — Subtract  the  length  of  the  segment,  i  c,  from  the 
half  length,  t  e,  of  the  spindle ;  double  the  remainder,  and  as- 
certain the  contents  of  a  middle  frustrum  of  this  length. 

Subtract  the  result  from  the  contents  of  the  whole  spindle, 
and  half  the  remainder  will  give  the  contents  of  the  segment 
required.* 

Or,  C—c-r-  2  =  S,  C  and  c  representing  the  contents  of  the  spin- 
dle and  middle  frustrum. 

Fig.  112. 


\   \    / 


o 


Example. — The  length  of  a  circular  spindle,  i  a,  Fig.  112, 
is  14.14213,  the  central  distance,  o  e,  is  7.07107,  the  radius 
of  the  arc,  o  a,  is  10,  and  the  length  of  the  segment,  t  c,  is 
3.53553  inches;  what  are  its  contents'? 

— '— 3. 53553  X  2 =7. 071 07 =doub!e  the  remainder,  of  the  length 

of  the  segment  subtracted  from  half  the  length  of  the  spindle  =  length  of  the, 
middle  frustrum . 

Note. — The  area  of  the  revolving  or  generating  segment  of  the  whole 
spindle  is  28.54  inches,  and  that  of  the  middle  frustrum  is  19.25. 

The  contents  of  the  whole  spindle  is 212.9628  cubic  in. 

middle  frustrum  is 162.8982     "     " 

Hence 50.0646^2  = 

25.0323= Me  contents  required. 

*  This  rule  is  applicable  to  the  segment  of  any  spindle  or  any  conoid, 
the  volume  of  the  figure  and  frustrum  being  first  obtained. 


192 


MENSURATION   OF    SOLIDS. 


Centre  of  Gravity.* 

^(r3+a2)(^-Za)-i(^-^)+|a[a3-(^-P)f] 
(r>+a*X9-l)-^-t3)-2aS 


of  spindle. 

When  g=b  o. 
l—o  c. 

r— radius  of  circle,  a  d. 
a=a  o  =  V  r2— #2. 
S=area  of  the  semi-zone,   / 
efff  h.  / 


= distance  from  centre 


Cycloidal  Spindle.^ 

To  ascertain  the  Contents  of  a  Cycloidal  Spindle,  Fig.  113. 

Rule. — Multiply  the  product  of  the  square  of  twice  the  di- 
ameter of  the  generating  circle,  a  b  c,  and  3.927  by  its  circum- 
ference, and  this  product  divided  by  8  will  give  the  contents 
required. 


Or, 


2d  x  3.927x^x3.1416 
8 


S,  d  representing  the  diameter 


of  the  circle,  a  half  width  of  the  spindle. 
Fig.  113. 


\ 
/c 


Example. — The  diameter  of  the  generating  circle,  a  b  c,  of 
a  cycloid,  Fig.  113,  is  10  inches  ;  what  are  the  contents  of  the 
spindle,  d  e  ? 

*  By  Professor  A.  E.  Church,  U.  S.  M.  A. 

t  The  contents  of  a  cycloidal  spindle  are  equal  to  §  of  its  circumscrib- 
ing cylinder. 


MENSURATION   OF    SOLIDS.  193 


10X2X3.927  =  1570. 8 ^product  of  twice  the  diameter  squared  and 
3.927.         

1570.8  X  10X3. 1416-f-8 =6168.5316  cubic  inches  =product  of  the  pre- 
ceding product  and  the  circumference  divided  by  8  =  result  required. 

Ex.  2.  The  diameter  of  the  generating  circle  of  a  cycloid  is 
5G.5  inches ;  what  are  the  contents  of  its  spindle  in  cubic  feet  ? 

Ans.  64:3.9292  cubic  feet. 
Ex.  3.  The  diameter  of  the  generating  circle  of  a  cycloid  is 
6  feet ;  what  are  the  contents  of  its  spindle  in  cubic  feet  ? 

Ans.  1332.4028  cubic  feet 

Elliptic  /Spindle. 

To  ascertain  the  Contents  of  an  Elliptic  Spindle,  Fig.  114. 
Rule. — To  the  square  of  its  diameter,  c  d,  add  the  square 
of  twice  the  diameter  ef  at  J  of  its  length ;  multiply  the  sum 
by  the  length,  a  b,  the  product  by  .1309,  and  it  will  give  the 
contents  required. 

Note. — For  all  such  solids,  this  rule  is  exact  when  the  body  is  form- 
ed by  a  conic  section,  or  a  part  of  it,  revolving  about  the  axis  of  the 
section,  and  will  always  be  very  near  when  the  figure  revolves  about 
another  line. 

Or,  d2-\-2d'2  x^X«1309=:S,  d  and  <t  representing  the  diam- 
eters as  above. 

Fig.  114. 


Example. — The  length  of  an  elliptic  spindle,  a  b,  Fig.  114, 
is  75  inches,  its  diameter  c  d  35,  and  the  diameter  efat  £  of 
its  length  25  ;  what  are  its  contents? 

I 


194 


MENSURATION    OF    SOLIDS. 


35s +25  X  2 =3725=  sum  of  squares  of  diameter  of  spindle  and  of  twice 
its  diameter  at  £  of  its  length. 

3725  X  75 =279375 =above  sum  X  the  length  of  the  spindle. 
Then  279375  X.  1309 =36570.1875 =result  required. 

Ex.  2.  The  length  of  an  elliptic  spindle  is  100  inches,  its 
diameter  20,  and  the  diameter  at  J  of  its  length  15  ;  what  are 
its  contents?  Ans.  17017  cubic  inches. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Contents  of  the  Middle  Frustrum  or  a  Zone  of 
an  Elliptic  Spindle,  Fig.  115. 

Rule. — Add  together  the  squares  of  the  greatest  and  least 
diameters,  a  b,  c  d,  and  the  square  of  double  the  diameter  g  h 
in  the  middle  between  the  two ;  multiply  the  sum  by  the 
length,  e  f,  the  product  by  .1309,  and  it  will  give  the  con- 
tents required.* 

Or,  d2+<r2+(2xcO2x/x.l309=S,  d,d',  and  d"  repre- 
senting the  different  diameters. 

Fig.  115. 


Example. — The  greatest  and  least  diameters,  a  b  and  c  d,  of 
the  frustrum  of  an  elliptic  spindle,  Fig.  115,  are  68  and  50  inches, 
its  middle  diameter,  g  h,  60,  and  its  length,  ef  75 ;  what  are  its 
contents  ! 


G82  +  502  +  G0x2=21524=swm  of  squares  of  greatest  and  least  diam- 
eters and  of  double  the  middle  diameter. 

21524  X  75  X.  1309=211311. 87=product  of  above  sum,  the  length,  and 
.  1 309 = the  result  required. 

*  For  all  such  solids,  this  rule  is  exact  when  the  body  is  formed  by 
a  conic  section,  or  a  part  of  it,  revolving  about  the  axis  of  the  section, 
and  will  always  be  very  near  when  the  figure  revolves  about  another 
line. 


MENSURATION    OF   SOLIDS.  195 

Ex.  2.  The  greatest  and  least  diameters  of  the  zone  of  an 
elliptic  spindle  are  20  and  5  inches,  its  middle  diameter  16, 
and  its  length  42 ;  what  are  its  contents  ? 

Ans.  7966.312  cubic  inches. 

Ex.  3.  The  greatest,  middle,  and  least  diameters  of  the  zone 
of  an  elliptic  spindle  are  25,  23.5,  and  20  inches,  and  its  length 
42.5;  what  are  its  contents'?      Ans.  17991.55  cubic  inches. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Contents  of  a  Segment  of  an  Elliptic  Spindle, 
Fig.  116. 

Rule. — Add  together  the  square  of  the  diameter  of  the 
base,  c  d,  of  the  segment,  and  the  square  of  double  the  diam- 
eter g  h,  in  the  middle  between  the  base  and  vertex ;  multiply 
the  sum  by  the  length,  o  e,  of  the  segment,  the  product  by 
.1309,  and  it  will  give  the  contents  required.* 

Or,  rf2+2  d//2x/x.l309  =  S,  d  and  d"  representing  the  di- 
ameters. 

Fig.  116. 


--> 


d      — 

Example. — The  diameters,  c  d  and  g  h,  of  the  segment  of 
an  elliptic  spindle,  Fig.  116,  are  20  and  12  inches,  and  the 
length,  o  e,  is  16  inches ;  what  are  its  contents? 


20a  +  12x2  =  976=£ttfrc  of  squares  of  diameter  at  base  and  of  double  the 
diameter  in  the  middle. 

976  X  16  X.  1309 =2044.134  =product  of  above  sum,  the  length  of  the 
segment,  and  .1309  =  <^e  result  required. 

Ex.  2.  The  diameters  of  the  segment  of  an  elliptic  spindle 
are  25  and  20  inches,  and  the  length  of  it  is  42.5  inches ; 
what  are  its  contents'?  Ans.  12378.231  cubic  inches. 

Centre  of  Gravity.  At  two  thirds  of  the  length,  measuring 
from  the  end. 

*  See  note  at  bottom  of  page  194. 


196  MENSURATION    OF    SOLIDS. 

Parabolic  Spindle. 

To  ascertain  the  Contents  of  a  Parabolic  Spindle,  Fig.  117. 

Rule  1. — Multiply  the  square  of  the  diameter,  a  b,  by  the 
length,  c  d,  the  product  by  .41888,*  and  it  will  give  the  con- 
tents required. 

Or,  d2xl  XA1888=S. 

Rule  2. — To  the  square  of  its  diameter  add  the  square  of 
twice  the  diameter  at  \  of  its  length ;  multiply  the  sum  by 
the  length,  the  product  by  .1309,  and  it  will  give  the  contents 
required.! 

Or,  d2+2d'2  x  lx  .1309  =  S,  d  and  d/  representing  the  diam- 
eters as  above. 

Fig.  117.  c 


Example. — The  diameter  of  a  parabolic  spindle,  a  b,  Fig. 
117,  is  40  inches,  and  its  length,  c  d,  20 ;  what  are  its  con- 
tents ? 

402  X  20 =32000= square  of  diameter  X  the  length. 
Then  32000 X. 41888  =  13404. 16 =a&ore  product X  A1888  =result  re- 
quired. 

Again,  If  the  middle  diameter  at  one  fourth  of  its  length  is  29.  G5, 
then,  by  Rule  2, 


40-+29.65x2x20X.1309  =  13394.97=resuftre£i«Vee/. 
Ex.  2.  The  length  of  a  parabolic  spindle  is  15.75  feet,  and 
its  diameter  is  3  feet ;  what  are  its  contents  ? 

Ans.  59.376  cubic  feet. 

*  T8sof.7854. 

f  See  note  at  bottom  of  page  194. 


MENSURATION    OF    SOLIDS.  197 

Ex.  3.  The  length  of  a  parabolic  spindle  is  40  feet,  and 
its  diameter  is  80  feet ;  what  are  its  contents  in  cubic  feet  t 

Ans.  107233.28  cubic  feet.    ■ 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Contents  of  the  Middle  Frustrum  of  a  Parabolic 
Spindle,  Fig.  118. 

Rule  1. — Add  together  8  times  the  square  of  the  greatest 
diameter,  a  b,  3  times  the  square  of  the  least  diameter,  e  f,  and 
4  times  the  product  of  these  two  diameters ;  multiply  the  sum 
by  the  length,  c  d,  the  product  by  .05236,  and  it  will  give  the 
contents  required. 

Or,  d2x84-^^3+^xd7x4x^X.05236  =  S. 

Rule  2. — Add  together  the  squares"of  the  greatest  and  least 
diameters,  a  b,  ef,  and  the  square  of  double  the  diameter  in 
the  middle  between  the  two ;  multiply  the  sum  by  the  length, 
c  d,  the  product  by  .1309,  and  it  will  give  the  contents  re- 
quired. 

Or,  cZ2  +  d/2+(2O2xJx.l309=S,  d"  representing  the  di- 
ameter between  the  two. 


Fig.  118. 


c  **«•«. 


•— - 

Example. — The  middle  frustrum  of  a  parabolic  spindle, 
Fig.  118,  has  diameters,  a  b  and  ef  of  40  and  30  inches,  and 
its  length,  c  d,  is  10  inches ;  what  are  its  contents'? 

402  X  8+302  X3-H0x30x4=20300=^e  sum  of  8  times  the  square  of 
ihc  greatest  diameter,  3  times  the  square  of  the  least  diameter,  and  A;  times 
the  product  of  these  diameters. 

20300 X  10 X .05236  =  10G29.08=resuft  required. 


198  MENSURATION   OP   SOLIDS. 

Ex.  2.  The  middle  frustrum  of  a  parabolic  spindle  has  di- 
ameters of  20  and  15  feet,  and  its  length  5  feet ;  what  are 
its  contents?  Ans.  1328.635  cubic  feet. 

Ex.  3.  The  middle  frustrum  of  a  parabolic  spindle  has  di- 
ameters of  80  and  56.5  feet,  and  its  length  is  20- feet;  what 
are  its  contents'?  Ans.  82578.789  cubic  feet. 

Centre  cf  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Contents  of  a  Segment  of  a  Parabolic  Spindle, 
Fig.  119. 

Rule. — Add  together  the  square  of  the  diameter  of  the 
base,  e  f,  of  the  segment,  and  the  square  of  double  the  diam- 
eter, g  h,  in  the  middle  between  the  base  and  vertex ;  multiply 
the  sum  by  the  length,  c  d,  of  the  segment,  the  product  by 
.1309,  and  it  will  give  the  contents  required. 

Or,  d2+2  dr2x/X.1309=S. 


Example. — The  segment  of  a  parabolic  spindle,  Pig.  119, 
has  diameters,  e  f  and  g  h,  of  15  and  8.75  inches,  and  the 
length,  c  d,  is  2.5  inches ;  what  are  its  contents  ? 


152  +  8.75x  2=531. 25=sum  of  square  of  base  and  of  double  the  diam- 
eter in  the  middle  of  the  segment. 

531.25 X 2.5 X.130S =173.852  =product  of  above  sum,  length  of  seg- 
ment, and  .1309  =the  result  required. 

Ex.  2.  The  segment  of  a  parabolic  spindle  has  diameters 
of  30  and  20  inches,  and  the  length  of  it  is  30  inches ;  what 
are  its  contents  in  cubic  feet?  Ans.  5.6814  cubic  feet. 


MENSURATION   OF   SOLIDS.  199 

Ex.  3.  The  segment  of  a  parabolic  spindle  has  diameters 
of  56.5  and  40  feet,  and  the  length  of  it  is  10  feet ;  what 
are  its  contents  in  cubic  feet  ? 

Ans.  12556.255  cubic  feet 

Ex.  4.  The  segment  of  a  parabolic  spindle  has  diameters 
of  28  and  20  feet,  and  the  length  of  it  is  20  feet ;  what  are 
its  contents?  Ans.  6241.312  cubic  feet. 

Ex.  5.  The  segment  of  a  parabolic  spindle  has  diameters 
of  42  and  30  feet,  and  the  length  of  it  is  60  feet ;  what  are 
its  contents'?  Ans.  42128.856  cubic  feet. 

Centre  of  Gravity* 

b6     d6 

u      "ap  (bi-di)+2a2p2(b2-d2) 

—  — —distance  from  centre 

h  J^J^L^-d^+latp^b-d) 
5       5         o 

of  spindle,  a  representing  semi-diameter  of  spindle,  b  the  half 

length,  d  the  distance  of  the  base  of  the  segment  from  the  centre 

b2 
of  the  spindle,  andp  one  half  parameter,  which  is  equal  to  — . 


Illustration  of  rules,  p.  196,  197,  and  198 : 

Solidity  of  spindle  (Ex.  3,  p.  197)  by  rule  2, 
80  feet  in  diameter  by  40  feet  in  length, 
the  diameter  at  £  of  its  length  being  56.5 
feet =  100368.884  cubicfeet. 

Solidity  of  middle  frustrum  (Ex.  3,  p.  198)  by  rule  2, 
20  feet  in  length,  the  diameter  at  £  of  its  length 
being  69.25  feet \ =  75331.641 

Solidity  of  segment  (Ex.  3,  p.  199)  10  feet  in  length 

=  12556.255,  which  x  2,  for  two  end  segments =  25112.510 

100444.151 

Solidity  of  spindle  as  above =  100368.884 

Difference,  arising  from  the  impracticability  of  ob- 
taining the  middle  diameters  of  the  frustrum  and 
segment,  from  a  figure  of  so  minute  a  scale  as 
that  of  the  example  taken  for  illustration =         75.267 

*  By  Professor  A.  E.  Church,  U.  S.  M.  A. 


200  MENSURATION   OF    SOLIDS. 

Hyperbolic  Spindle. 

To  ascertain  the  Contents  of  a  Hyperbolic  Spindle,  Fig.  120. 

Rule. — To  the  square  of  its  diameter,  c  d,  add  the  square 

of  double  the  diameter,  e  f  at  ^  of  its  length ;  multiply  the 

sum  by  the  length,  a  b,  the  product  by  .1309,  and  it  will  give 

the  contents  required* 

Or,  d2-f-2d/2x£x.l309:=S,  d  and  d'  representing  the  diam- 
eters as  above. 

Fig.  120.  a 


b 
Example. — The  length,  a  b,  Fig.  120,  of  a  hyperbolic  spin- 
dle is  100  inches,  and  its  diameters,  c  d  and  ef  are  150  and 


1503  +  110  X  2  X 100  =  7090000  =product  of  the  sum  of  the  squares  of  the 
greatest  diameter  and  of  twice  the  diameter  at  £  of  the  length  of  the  spin- 
dle and  the  length. 

Then,  7090000  X.  1309  =928081  =  the  result  required. 

Ex.  2.  The  length  of  a  hyperbolic  spindle  is  120  inches, 
and  its  diameters  in  the  middle  and  at  J  of  its  length  are  100 
and  80  inches ;  what  are  its  contents  % 

Ans.  323.614  cubic  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Contents  of  the  Middle  Frustrum  of  a  Hyper- 
bolic Spindle,  Fig.  121. 
Rule. — Add  together  the  squares  of  the  greatest  and  least 
diameters,  a  b,c  d,  and  the  square  of  double  the  diameter,  g  h, 
in  the  middle  between  the  two ;  multiply  this  sum  by  the 
*  See  note  at  bottom  of  page  194. 


MENSURATION    OF    SOLIDS.  201 

length,  e  f  the  product  by  .1309,  and  it  will  give  the  contents 
required.* 

Or,  rf2+^2+(2x^//)2x/x.l309  =  S. 
Fig.  121.  ....- ; 


^b 


m 

^ 1. — ..-^ 


Example. — The  diameters  a  b  and  c  d,  of  the  middle  frus- 
trum of  a  hyperbolic  spindle,  Fig.  121,  are  150  and  110  inches, 
the  diameter  g  h,  140  inches,  and  the  length,  e  f,  50;  what 
are  its  contents  I 


1502  +  1102  +  14:0x2  =  113000=swm  of  squares  of  greatest  and  least  di- 
ameters and  of  double  the  middle  diameter. 

113000  X  50  X  .1309  =  739585  =product  of  above  sum  X  the  length  X  .1309 
=7-esult  required. 

Ex.  2.  The  diameters  of  the  middle  frustrum  of  a  hy- 
perbolic spindle  are  16  and  10  inches,  the  diameter  at  J 
of  its  length  is  13.5,  and  the  length  of  it  is  l6 ;  what  are 
its  contents  in  cubic  inches'?       Ans.  1420.265  cubic  inches. 

Ex.  3.  The  diameters  of  the  middle  frustrum  of  a  hyper- 
bolic spindle  are  16  and  12  feet,  the  diameter  at  J  of  its  length 
14.5,  and  the  length  of  it  20  feet ;  what  are  its  contents  ? 

Ans.  3248.938  cubic  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 

To  ascertain  the  Contents  of  a  Segment  of  a  Hyperbolic  Spindle, 
Fig.  122. 
Rule. — Add  together  the  square  of  the  diameter  of  the 
base,  e  f,  of  the  segment,  and  the  square  of  double  the  diam- 
eter, g  h,  in  the  middle  between  the  base  and  vertex ;  multiply 
the  sum  by  the  length,  a  b,  of  the  segment,  the  product  by 
.1309,  and  it  will  give  the  contents  required. 
*  See  note  at  bottom  of  page  194. 
12 


202  MENSURATION   OF    SOLIDS. 

Or,  d2+<f/2x*X.1309  =  S,  d  and  d" representing  the  diam- 
eters. 

Fig.  122.  a 


Example. — The  segment  of  a  hyperbolic  spindle,  Fig.  122, 
has  diameters,  e  f  and  g  h,  of  110  and  65  inches,  and  its 
length,  a  b,  25  ;  what  are  its  contents  ? 


1102  +  65x2=29000=sttm  of  squares  of  diameter  of  base  and  of  double 
the  middle  diameter. 

29000  X  25  X  .1309 =94902.5  =product  of  above  sum  X  the  length  X  .1309 
=result  required. 

Ex.  2.  The  segment  of  a  hyperbolic  spindle  has  diameters 
of  8  inches  at  its  base  and  6  at  half  its  length,  its  length  be- 
ing 10 ;  what  are  its  contents? 

Ans.  272.272  cubic  inches. 
Ex.  3.  The  segment  of  a  hyperbolic  spindle  has  diameters 
of  50  and  31  inches  at  its  base  and  half  its  length,  its  length 
being  25  ;  what  are  its  contents  ? 

Ans.  20760.74  cubic  inches. 

Centre  of  Gravity.  At  J  of  the  height,  measured  from  the 
vertex. 


ellipsoid,  paraboloid,  and  hyperboloid  of  revolution* 
(conoids). 

Definition.  Figures  like  to  a  cone,  described  by  the  revolu- 
tion of  a  conic  section  around  and  at  a  right  angle  to  the  plane  of 
their  fixed  axes. 

*  These  figures  have  been  known  as  conoids.  For  the  definition  of 
a  conoid,  see  Conic  Sections,  page 


MENSURATION   OP   SOLIDS. 


203 


Ellipsoid  of  Revolution  {Spheroid). 

An  ellipsoid  of  revolution  is  a  semi-spheroid.     (See  p. 
180-184.) 


Paraboloid  of  Revolution* 

To   ascertain  '  the    Contents   of  a  Paraboloid   of  Revolution, 
Fig.  123. 
Rule. — Multiply  the  area  of  the  base,  a  b,  by  half  the  alti- 
tude, d  c,  and  the  product  will  give  the  contents  required. 

Note. — This  rule  will  hold  for  any  segment  of  the  paraboloid,  whether 
the  base  be  perpendicular  or  oblique  to  the  axis  of  the  solid. 

or,  «x|==a 

Fig.  123. 


Example. — The  diameter,  a  b,  of  the  base  of  a  paraboloid 
of  revolution,  Fig.  123,  is  20  inches,  and  its  height,  d  c,  20 
inches ;  what  are  its  contents  ? 

Area  of  20  inches  diameter  of  base  =314.16. 

20 
314.16  X  — =3141. 6=area  of  base  X  half  the  height=result  required. 

Ex.  2.  The  diameter  of  the  base  of  a  paraboloid  of  revolu- 
tion is  11.5  inches,  and  its  height  7  ;  what  are  its  contents? 

Ans.  363.5411  cubic  inches. 

*  The  contents  of  a  paraboloid  of  revolution  are=i  of  its  circum- 
scribing cylinder. 


204  MENSURATION    OP    SOLIDS. 

Ex.  3.  The  diameter  of  the  base  of  a  paraboloid  of  revolu- 
tion is  11  feet  3  inches,  and  its  height  8  feet;  what  are  its 
contents  in  cubic  feet?  Ans.  397.608  cubic  feet. 

Centre  of  Gravity.  At  two  thirds  of  the  height,  measured 
from  the  vertex. 

Frustrum  of  a  Paraboloid  of  Revolution. 

To  ascertain  the  Contents  of  a  Frustrum  of  a  Paraboloid  of  Hev-> 
olution,  Fig.  124. 

Rule. — Multiply  the  sum  of  the  squares  of  the  diameters, 
a  b  and  d  c,  by  the  height,  e  f,  of  the  frustrum,  this  product 
by  .3927,  and  it  will  give  the  contents  nearly. 

Or,  d2-fd/2x^X.3927=S. 
Fig.  124. 


Example. — The  diameters,  a  b  and  d  c,  of  the  base  and 
vertex  of  the  frustrum  of  a  paraboloid  of  revolution,  Fig.  124, 
are  20  and  11.5  inches,  and  its  height,  ef  12.6;  what  are 
its  contents  ? 

202  +  11.52=5B2.25=sum  of  squares  of  the  diameters. 

532.25  X  12.6  X. 3927=2033.5837 ^product  of  above  sum,  the  height, 
and  .3927=^6  result-  required. 

Ex.  2.  The  diameters  of  the  frustrum  of  a  paraboloid  of 
revolution  are  30  and  58  inches,  and  the  height  18 ;  what  are 
its  contents  in  cubic  feet?  Ans.  17.4424  cubic  feet. 

Ex.  3.  The  diameters  of  the  frustrum  of  a  paraboloid  of 

revolution  are  48  and  60  inches,  and  the  height  18 ;  what  are 

its  contents  in  cubic  feet?  Ans.  24.151  cubic  feet. 

2R2-f-r2 
Centre  of  Gravity.     From  the  vertex  JA  2  ,  R  and  r 

representing  radii  of  base  and  diameter,  and  h  the  height. 


MENSURATION   OF    SOLIDS.  205 


Segment  of  a  Paraboloid  of  Revolution. 

To  ascertain  the  Contents  of  the  Segment  of  a  Paraboloid  of 
Revolution,  Fig.  125.    * 

Rule. — Multiply  the  area  of  the  base,  a  b,  by  half  the  alti- 
tude, e  f  and  the  product  will  give  the  contents  required. 

Note. — This  rule  will  hold  for  any  segment  of  the  paraboloid,  whether 
the  base  be  perpendicular  or  oblique  to  the  axis  of  the  solid. 

^  h      ^ 

Ortax^=S. 

Fig.  125. 


Example. — The  diameter,  a  b,  of  the  base  of  a  segment  of 
a  paraboloid  of  revolution,  Fig.  125,  is  11.5  inches,  and  its 
height,  e  /,  is  7.4  ;  what  are  its  contents  ? 

■* 
Area  of  11.5  inches  diameter  of  base  =  103.869. 

7.4 
103.869  X— =384.315  —arm  of  base  X  half  the  height=the  result  re- 

quired.  ... 

Ex.  2.  The  diameter  of  the  base  of  a  segment  of  a  parabo- 
loid of  revolution  is  30  inches,  and  its  height  25  ;  what  are 
its  contents  in  cubic  inches  ? 

Ans.  8835.73  cubic  inches. 

Ex.  3.  The  diameter  of  the  base  of  a  segment  of  a  parabo- 
loid of  revolution  is  48  inches,  and  the  height  15 ;  what  are 
its  contents  in  cubic  feet?  Ans.  7.854  cubic  feet. 

Centre  of  Gravity.  At  two  thirds  of  the  height,  measured 
from  the  vertex. 


206  MENSURATION   OF   SOLIDS. 


Hyperboloid  of  Revolution. 

To  ascertain  the  Contents  of  a  Hyperboloid  of  Revolution,  Fig. 

126. 

Rule. — To  the  square  of  the  radius  of  the  base,  a  b,  add 
the  square  of  the  middle  diameter,  c  d;  multiply  this  sum  by 
the  height,  ef  the  product  by  .5236,  and  it  will  give  the  con- 
tents required. 

Or,  r2-(-cZx/«X.5236r=S,  d  representing  the  middle  diam- 
eter. 


Fig.  126, 


Example. — The  base,  a  b,  of  a  hyperboloid  of  revolution, 
Fig.  126,  is  80  inches,  the  middle  diameter,  c  d,  66,  and  the 
height,  ef  60  ;  what  are  its  contents  ? 


8(M-2  +  662=5956=s«m  of  square  of  radius  of  base  and  middle  diam- 
eter. 

5956  X  50=297800  X  .5236  =  155928. 08  =product  of  above  sum,  the 
height,  and  .52S6=result  required. 

Ex.  2.  The  base  of  a  hyperboloid  of  revolution  is  20  inches, 
its  middle  diameter  is  16,  and  its  height  20 ;  what  are  its  con- 
tents !  Ans.  3728.032  cubic  inches. 

Ex.  3.  The  base  of  a  hyperboloid  of  revolution  is  104 
inches,  the  diameter  at  half  its  height  is  68,  and  the  height  of 
it  is  50  inches;  what  are  its  contents  in  cubic  feet? 

Ans.  111.0226  cubic  feet 

Centre  of  Gravity.     Distance  from  the  vertex,  x  h, 

b  representing  the  base,  and  h  the  height  of  the  figure. 


MENSURATION   OF    SOLIDS.  207 


Frustrum  of  a  Hyperboloid  of  Revolution, 

To  ascertain  the  Contents  of  the  Frustrum  of  a  Hyperboloid  of 
Revolution,  Fig.  127- 

Rule. — Add  together  the  squares  of  the  greatest  and  least 
semi-diameters,  e  b  and  *  d,  and  the  square  of  the  whole  di- 
ameter, g  h,  in  the  middle  of  the  two ;  multiply  this  sum  by 
the  height,  i  e,  the  product  by  .5236,  and  it  will  give  the  con- 
tents required. 

Or,  d2+d'2-{-d/'2xhx.5236=S,  d,  d',  and  d"  represent- 
ing  the  several  diameters. 

Fig.  127.  f 


Example. — The  frustrum  of  a  hyperboloid  of  revolution, 
Fig.  127,  is  in  height,  e  i,  50  inches,  the  diameters  of  the  great- 
er and  less  ends,  a  b  and  c  d,  are  110  and  42,  and  that  of  the 
middle  diameter,  g  h,  is  80 ;  what  are  the  contents  % 

110-^2=55  and  42-f-2=21.  Hence,  552+212+802=9866=sw7w  of 
the  squares  of  the  semi-diameters  of  the  ends  and  that  of  the  whole  diame- 
ter in  the  middle. 

9866  X  50=493300  X  .5236=258291.88  cubic  inches. 

Ex.  2.  The  height  of  a  frustrum  of  a  hyperboloid  of  revolu- 
tion is  1  foot,  the  greatest  and  least  diameters  10  and  6  inch- 
es, and  the  middle  diameter  8.5  inches;  what  are  its  con- 
tents in  cubic  feet?  Ans.  .38633  cubic  foot. 

Ex.  3.  The  height  of  the  frustrum  of  a  hyperboloid  of  rev- 
olution is  10  inches,  the  radii  of  the  ends  21  and  1  inches,  and 
the  middle  diameter  25  inches ;  what  are  its  contents  in  cubic 
feet?  Ans.  3.2331  cubic  feet. 


208  MENSURATION    OF    SOLIDS. 

Centre  of  Gravity. 

_  (d+d>)(2a2-d/a4-d2)      »  ,  .  *-«         . 

x     ^   ,       ,,, -7 — ^ -^distance  from  centre  of  the  base  of 

*     3a2— d't+d'd+d2  J  J  J 

the  figure,  a  representing  the  semi-transverse  axis,  or  distance  from 

centre  of  the  curve  to  vertex  of  figure,  f;  d  and  d/  the  distances 

from  the  centre  of  the  curve  to  the  centre  of  the  less  and  greater 

diameter  of  the  frustrum. 

Segment  of  a  Hyperboloid  of  Revolution. 

To  ascertain  the  Contents  of  the  Segment  of  a  Hyperboloid  of 
Revolution,  Fig.  128. 

Eule. — To  the  square  of  the  radius  of  the  base,  a  e,  add 
the  square  of  the  middle  diameter,  c  d ;  multiply  this  sum  by 
the  height,  e  f,  the  product  by  .5236,  and  it  will  give  the  con- 
tents required. 

Or,  r2 -\-d2  X  h  x  .5236  ==  S,  r  representing  radius  of  base.     • 

Fig.  128.  f 


Example. — The  radius,  a  e,  of  the  base  of  a  segment  of  a 
hyperboloid  of  revolution,  Fig.  128,  is  21  incnes,  its  middle  di- 
ameter, c  d,  is  30,  and  its  height,  e  f,  15  ;  what  are  its  con- 
tents 1 

21 2+302X  15=20115=<Ae  product  of  the  sum  of  the  squares  of  the  ra- 
dius of  the  base  and  the  middle  diameter  multiplied  by  the  height. 
20115  X  .5236  =  10532.214=resM&  required. 

Ex.  2.  The  radius  of  the  base  of  a  segment  of  a  hyperboloid 
of  revolution  is  55  inches,  its  middle  diameter  70,  and  its 
height  65  ;  what  are  its  contents? 

Ans.  269719.45  cubic  inches. 

Centre  of  Gravity.     (See  rule  for  hyperboloid,  page  206.) 


MENSURATION    OF    SOLIDS. 


209 


ANY   FIGURE    OF   REVOLUTION. 

To  ascertain  the  Contents  of  any  Figure  of  Revolution,  Fig.  129. 

Rule. — Multiply  the  area  of  the  generating  surface  by  the 
circumference  described  by  its  centre  of  gravity. 

Or,  ax%rxp=contents,  r  representing  radius  of  centre  of 
gravity. 

Fig.  129. 


HH 


Example. — If  the  generating  surface,  a  b  c  d,  of  the  cylin- 
der, b  e  df,  Fig.  129,  is  5  inches  in  width  and  10  in  height, 
then  will  a  b=5  and  b  d  =  10,  and  the  centre  of  gravity  will 
be  in  o,  the  radius  of  which  is  r  o— 5-r-2  =  2.5. 

Hence,  10x5  =  50 =area  of  generating  surface. 

50  X  2.5  X  2  X  3.1416 =785.4 =area  X  circumference  of  its  centre  of grav- 
ity =  the  contents  of  the  cylinder. 

Proof.  Volume  of  a  cylinder  10  inches  in  diameter  and  10  inches  in 
height.     10s X. 7851  =78.54,  and  78.54x10=785.4. 
'  Fig.  130. 


Ex.  2.  If  the  generating  surface  of  a  cone,  Fig.  130,  is  a  e 
=  10,  d  e=5,  then  will  a  d— 11.18,  and  the  area  of  the  tri- 
angle =  10  x  5-^2  =  25,  the  centre  of  gravity  of  which  is  in  o, 
and  o  r,  by  rule,  page  57,  =1.666. 

Hence,  25  X  1.606x2  X3. 1416 =261.8  =area  of  generating  surfaceX 
circumference  of  its  centre  of  gravity •  =  the  contents  of  the  cone. 


210  MENSURATION   OP   SOLIDS. 

Fig.  131. 


Ex.  3.  If  the  generating  surface  of  a  sphere,  Fig.  181,  is 

/102x.7854\ 
a  b  c,  and  ac=10,  ab  c  will  be  I - J  =39.27,  the  cen- 
tre of  gravity  of  which  is  in  o,  and  by  rule,  page  87,  or— 2.122. 

Hence,  39.27  X  2.122  x  2  X  3.1416  =523.6  =area  of  generating  surface 
X  circumference  of  its  centre  of  gravity  —  the  contents  of  the  sphere. 

To  ascertain  the  Contents  of  an  Irregular  Body. 

Rule. — Weigh  it  both  in  and  out  of  fresh  water,  and  note 
the  difference  in  pounds;  then,  as  62.5*  is  to  this  difference, 
so  is  1728f  to  the  number  of  cubic  inches  in  the  body. 

Or,  divide  the  difference  in  pounds  by  62.5,  and  the  quo- 
tient will  give  the  volume  in  cubic  feet. 

Note. — If  salt  water  is  to  be  used,  the  ascertained  weight  of  a  cubic 
foot  of  it,  or  64,  is  to  be  used  for  62.5. 

Example. — An  irregular  shaped  body  weighs  15  pounds  in 
water,  and  30  out ;  what  is  its  volume  in  cubic  inches  ? 

30  — 15  =  15=  difference  of  weights  in  and  out  of  water. 

62.5  :  15 : :  1728 :  414.72  =volume  of  the  body  in  cubic  inches. 

Or,  15^62.5  =  .24,  and  .24  X  1728 =414.72  =volume  of  the  body. 

Ex.  2.  An  irregular  body  weighs  187.5  pounds  in  water, 
and  250  out ;  what  is  its  volume?     Ans.  1728  cubic  inches. 

Ex.  3.  The  difference  in  weights  of  a  bronze  gun  in  and  out 
of  water  is  625  pounds ;  what  is  its  volume  in  cubic  feet? 

Ans.  10  cubic  feet 

*  The  weight  of  a  cubic  foot  of  fresh  water. 
•     f  The  number  of  inches  in  a  cubic  foot. 


MENSURATION  'OF   SOLIDS.  211 


PROMISCUOUS  EXAMPLES. 

1.  If  a  stone  measures  4  feet  6  inches  long,  2  feet  9  inches 
broad,  and  5  feet  4  inches  deep,  how  many  cubic  feet  does  it 
contain?  Ans.  66  feet. 

2.  The  dimensions  of  a  bushel  measure  are  18^  inches  in 
diameter,  and  8  inches  deep ;  what  should  be  the  dimensions 
of  a  measure  of  like  form  that  would  contain  8  bushels'? 

Ans.  37  inches  in  diameter,  and  16  inches  deep. 

3.  If  a  box,  of  plank  3.5  inches  thick,  is  4  feet  9  inches  in 
length,  3  feet  7  inches  in  breadth,  and  2  feet  11  inches  in 
height,  how  many  square  feet  did  it  require  to  make  the  box, 
how  many  cubic  feet  does  it  contain,  and  how  many  does  it 
measure  ? 

Ans.  70.208  square  feet,  29.167,  and  49.644  cubic  feet. 

4.  A  well  40  feet  in  depth  is  to  be  lined,  of  which  the  di- 
ameter is  6.5  feet,  the  thickness  of  the  wall  is  to  be  1.5  feet, 
leaving  the  inner  diameter  of  the  well  3.5  feet ;  how  many  cubic 
feet  of  stone  will  be  required?  Ans.  942.478  cubic  feet. 

5.  How  many  bricks,  exclusive  of  mortar,  8  inches  long, 
4  inches  wide,  and  2  inches  thick,  will  it  take  to  build  a  wall 
40  feet  long,  20  feet  high,  and  2  feet  thick  ! 

Ans.  43,200  bricks. 

6.  How  many  bricks,  exclusive  of  mortar,  will  it  take  to 
build  the  walls  of  a  house  which  is  80  feet  long,  40  feet  wide,  and 
25  feet  high,  the  walls  to  be  12  inches  thick,  the  bricks  being 
8  inches  long,  4  broad,  and  2  thick?     Ans.  159,300  bricks. 

7.  How  many  bricks  will  it  require  to  construct  the  walls 
of  a  house  64  feet  long,  32  feet  wide,  and  28  feet  high ;  the 
walls  are  to  be  1  foot  4  inches  thick,  and  there  are  to  be  three 
doors  7  feet  4  inches  high,  and  3  feet  8  inches  wide ;  also  14 
windows  3  feet  wide  and  6  feet  high,  and  16  windows  2  feet 
8  inches  wide  and  5  feet  8  inches  high :  each  brick  is  to  be  8 
inches  long,  4  inches  wide,  and  2  inches  thick  ? 

Ans.  171,990  bricks. 


212  MENSURATION    OF    SOLIDS.        » 

8.  If  a  garden  100  feet  long  and  80  feet  wide  is  to  be  in- 
closed with  a  ditch  4  feet  wide,  how  deep  must  it  be  dug  that 
the  soil  taken  from  it  may  raise  the  surface  one  foot  ? 

Aiis.  5.319  feet. 

9.  If  a  man  dig  a  small  square  cellar,  which  will  measure 
6  feet  each  way,  in  one  day,  how  long  would  it  take  him  to 
dig  a  similar  one  that  measured  10  feet  each  way? 

Ans.  4.629  days. 

10.  If  a  lead  pipe  f  of  an  inch  in  diameter  will  fill  a  cis- 
tern in  3  hours,  what  should  be  its  diameter  to  fill  it  in  2 
hours'?  Ans.  .918  inch. 

11.  What  are  the  contents  of  a  stick  of  round  timber  20 
feet  long,  the  diameter  at  the  larger  end  being  12  inches, 
and  at  the  smaller  end  6  inches  ? 

Ans.  10.9126  cubic  feet. 

12.  Required  the  volume  of  Bunker  Hill  Monument,  the 
height  of  which  is  220  feet,  by  30  feet  square  at  its  base,  and 
15  feet  at  its*  vertex.  Ans.  115500  cubic  feet. 

13.  What  are  the  contents  of  a  spherical  segment  3  feet  in 
height,  cut  from  a  sphere  10  feet  in  diameter? 

Ans.  113.0976  cubic  feet. 

14.  The  largest  of  the  Egyptian  pyramids  is  square  at  its 
base,  and  measures  693  feet  on  a  side ;  its  height  is  500  feet. 
Supposing  it  to  come  to  a  point  at  its  vertex,  what  would  be 
its  contents,  and  how  many  miles  in  length  of  wall,  5  feet  in 
height  and  2  feet  thick,  would  it  make  ? 

An      (80041500  cubic  feet. 

•■    (,1515.9375  miles  in  length. 

15.  What  are  the  contents  of  a  sphere,  the  diameter  of 
which  is  20  inches'?  Ans.  4188.8  cubic  inches. 

16.  What  is  the  weight  of  an  iron  spherical  shell  5  inches 
in  diameter,  the  thickness  of  the  metal  being  1  inch,  estimat- 
ing a  cubic  inch  of  iron  to  weigh  .25  of  a  pound  ? 

Ans.  12.8282  jtounds. 

17.  How  many  cubic  feet  of  water  are  there  in  a  pond  that 
measures  200  acres,  and  is  20  feet  deep  ! 

Am.  174240000  cubic  feet 


MENSURATION    OF    SOLIDS.  213 

18.  If  rain  was  to  fall  to  the  depth  of  3  inches  on  a  surface 
of  20000  square  acres,  what  would  be  the  number  of  hogs- 
heads of  water  fallen,  assuming  each  hogshead  to  contain  100 
gallons,  and  each  gallon  231  cubic  inches? 

A        <  16292571  hhds.  42  galls.  3  qts.  0  pts.  3.43  gills, 
US'  1  or  217800000  cubic  feet. 

19.  The  ditch  of  a  fortification  is  1000  feet  long,  9  feet 
deep,  20  feet  broad  at  bottom,  and  22  at  top ;  how  much  wa- 
ter will  fill  the  ditch,  allowing  231  cubic  inches  to  make  a 
gallon?  Ans.  1413818.1819  gallons. 

20.  What  must  be  the  height  of  a  bin  that  will  contain 
600  bushels,  its  length  being  8  feet  and  breadth  41* 

Ans.  23.333  feet. 

Note. — As  a  bushel  contains  very  nearly  one  fourth  more  than  a 
cubic  foot,  the  dimensions  of  a  bin,  etc.,  for  any  required  number  of 
bushels,  may  be  readily  found  by  adding  one  fourth  to  the  number  of 
bushels  ;  the  result  will  give  the  number  of  cubic  feet  the  bin  will  con- 
tain, or  that  may  be  required.  Therefore,  when  two  dimensions  of  a  bin, 
etc.,  are  given,  divide  the  number  of  cubic  feet  by  their  product,  and  the 
quotient  will  be  the  other  dimension. 

In  the  above  example,  then, 

600-^-4  =  150,  and  600  +  150  =  750  =  ^Ae  number  of' cubic  feet  the  bin  is 
to  contain. 

Then  750-r-8  X  4  =£3. 4375  feet  =  the  height  of  the  bin  required. 

21.  The  length  of  a  bin  is  4  feet,  its  breadth  5  feet  6  inch- 
es ;  what  must  its  height  be,  by  the  above  rule,  that  it  may 
contain  272  bushels?  Ans.  15  feet  5.454  inches. 

22.  There  are  1000  men  besieged  in  a  town  with  provis- 
ions for  5  weeks,  allowing  each  man  16  ounces  a  day ;  if  they 
are  re-enforced  by  500  more,  and  no  relief  can  be  received  till 
the  end  of  8  weeks,  how  many  ounces  must  be  given  daily  to 
each  man  ?  Ans.  6.G6  ounces. 

23.  An  officer  drew  up  his  company  in  a  square,  the  num- 
ber in  each  rank  being  equal ;  on  being  re-enforced  with  three 
times  his  first  number  of  men,  he  placed  them  all  in  the  same 
form,  and  then  the  number  in  each  rank  was  just  double  what 
it  was  at  first ;  he  was  again  re-enforced  with  three  times  his 

*  The  standard  United  States  bushel  is  2150.42  cubic  inches. 


214  MENSURATION   OF    SOLIDS. 

whole  number  of  men,  and,  after  placing  them  all  in  the  same 
form  as  at  first,  his  number  in  each  rank  was  40  men ;  how 
many  men  had  he  at  first?  Am.  100  men. 

24.  The  volume  of  a  sphere  is  381.7044  cubic  inches  ;  re- 
quired its  radius.  Ans.  4.5  inches. 

25.  If  an  iron  wire  TL  of  an  inch  in  diameter  will  sustain  a 
weight  of  450  pounds,  what  weight  might  be  sustained  by  a 
wire  an  inch  in  diameter?  Ans.  45000  pounds. 

26.  The  edge  of  a  cube  is  36  inches ;  what  is  the  volume 
of  a  sphere  that  may  be  inscribed  within  it  ? 

Ans.  24429.0816  cubic  inches. 

27.  In  the  walls  of  Balbeck,  in  Turkey,  there  are  three 
stones  laid  end  to  end,  now  in  sight,  one  of  which  is  63  feet 
long,  12  feet  thick,  and  12  feet  broad ;  what  is  the  weight, 
supposing  its  specific  gravity  to  be  3  times  that  of  water? 

Ans.  759.375  tons. 

28.  If  two  men  carry  a  burden  of  200  pounds  suspend- 
ed near  the  middle  of  a  pole,  the  ends  of,  which  rest  on 
their  shoulders,  how  much  of  the  load  is  borne  by  each  man, 
it  hanging  6  inches  from  the  middle,  and  the  whole  length 
of  the  pole  being  4  feet  ? 

Ans.  125  pounds,  and  75  pounds. 

29.  A  joist  is  8  \  inches  deep  and  3^  broad;  what  will  be 
the  depth  of  a  beam  of  twice  the  contents  of  the  joist  that  is 
4f  inches  broad?  Ans.  12.526  inches. 

30.  Bunker  Hill  Monument  is  30  feet  square  at  its  base, 
15  feet  square  at  its  top,  and  its  height  is  220  feet;  from  the 
bottom  to  the  top,  through  its  centre,  is  a  cylindrical  opening 
15  feet  in  diameter  at  the  bottom  and  11  feet  at  the  top; 
how  many  cubic  feet  are  there  in  the  monument  ? 

Ans.  86068.444  cubic  feet. 

31.  If  a  bell  4  inches  in  height,  3  inches  in  diameter  (ex- 
ternal), and  i  of  an  inch  in  thickness,  weigh  2  pounds,  what 
should  be  the  dimensions  of  a  bell,  of  like  proportions,  that 
would  weigh  2000  pounds  % 

,       j 3  feet  4  inches  high,  2  feet  6  inches  diameter, 
'  \  and  2£  inches  thick. 


MENSURATION    OF   SOLIDS.  215 

32.  If  a  round  column  7  inches  in  diameter  has  a  capacity 
of  4  cubic  feet,  of  what  diameter  is  a  column  of  equal  length 
that  contains  10  times  as  much  ? 

Note. — The  contents  of  cylinders,  prisms,  parallelopipedons,  etc.,  of 
equal  altitudes,  are  to  each  other  as  the  squares  of  their  diameters  or 
like  sides.  The  same  rule  is  applicable  to  frustrums  of  a  cone  or  pyr- 
amid when  the  altitude  is  the  same  and  the  ends  proportional. 

Hence,  as  4  :  40,  or  as  1 :  10 : :  72 :  490  =  the  square  of  the  required  di- 
ameter, and  V  490 =22. 1359,  the  diameter  required. 

33.  A  frustrum  of  a  cone  is  12  inches  in  height,  and  the 
diameters  of  the  greater  and  smaller  ends  5  and  3  inches  re- 
spectively. Required  the  diameter  of  a  frustrum  of  the  same 
altitude  that  will  contain  3848.46  cubic  inches,  and  have  its 
diameters  in  the  same  proportion  as  the  smaller  one. 

Aixs.  The  greater  diameter  25,  and  less  diameter  15  inches. 

34.  There  is  a  fish  the  head  of  which  weighs  15  pounds, 
his  tail  weighs  as  much  as  his  head  and  half  as  much  as  his 
body,  and  his  body  weighs  as  much  as  his  head  and  tail.  Re- 
quired the  weight  of  the  fish.  Ans.  72  pounds. 

35.  A  certain  grocer  has  only  5  weights;  with  these  he 
can  weigh  any  quantity  by  pounds  from  1  to  121  pounds. 
Required  the  weights.         Ans.  1,  3,  9,  27,  and  81  pounds. 

36.  A  gentleman  has  a  bowling-green  300  feet  long  and 
200  feet  broad,  which  he  desires  to  raise  one  foot  higher  by 
means  of  earth  to  be  taken  from  a  ditch  that  is  to  go  around 
it;  to  what  depth  must  the  ditch  be  dug,  supposing  its 
breadth  to  be  8  feet?  Ans.  7  feet  3.21  inches. 

37.  Required  a  cylindrical  vessel  3  feet  in  depth  that  shall 
hold  twice  as  much  as  a  vessel  28  inches  deep  and  46  inches 
in  diameter  ;  what  must  be  its  diameter  *? 

Ans.  57.37  inches. 

38.  A  cubic  foot  of  brass  is  to  be  drawn  into  a  wire  of  ^_ 
of  an  inch  in  diameter ;  what  will  be  the  length  of  the  wire, 
assuming  there  is  to  be  no  loss  of  metal  in  the  operation  I 

Ans.  97784.5684  yards. 

39.  One  end  of  a  pile  of  wood  is  perpendicular  to  the  hori- 
zon, the  other  is  an  inclined  plane ;  the  length  of  the  pile  at 
the  bottom  is  64  feet,  at  the  top  50  feet,  in  height  12  feet,  and 


216  MENSURATION    OF    SOLIDS. 

the  length  of  the  wood  5  feet ;  required  the  number  of  cords 
it  contains?  Ans.  26  cords  92  feet. 

40.  If  a  vessel  of  war,  with  her  ordnance,  rigging,  and  ap- 
pointments, is  depressed  so  as  to  displace  50000  cubic  feet  of 
water,  what  is  the  weight  of  the  vessel"?* 

Ans.  1428.571  tons. 

41.  The  monument  erected  in  Babylon  by  Queen  Semira- 
mis  at  her  husband  Ninus's  tomb  is  said  to  have  been  one 
block  of  solid  marblef  in  the  form  of  a  square  pyramid,  the 
sides  of  the  base  being  20  feet,  and  the  height  of  the  monu- 
ment 150  feet ;  if  this  monument  had  been  sunk  in  the  Eu- 
phrates, what  weight  would  it  have  required  to  raise  its  apex 
to  the  surface  of  the  water  ?{  Ans.  2450000  pounds. 

42.  If  the  pyramid  described  in  the  last  example  were  di- 
vided into  three  equal  parts  by  planes  parallel  to  its  base, 
what  would  be  the  length  of  each  part,  measured  from  the 
top?  Ans.  104.0042,  27.0329,  and  18.9629  feet. 

43.  There  is  a  mill-hopper  in  the  form  of  a  square  pyra- 
mid, the  contents  of  which  are  13.5  feet,  the  side  of  its  great- 
er end  and  depth  are  in  the  proportion .  of  1  to  1.5  ;  but  one 
foot  has  to  be  cut  off  its  length  to  make  a  passage  for  the 
grain  from  the  hopper  to  -the  mill-stone.  Required  its  con- 
tents in  corn  measure"?  Ans.  10.446  bushels. 

44.  A  crucible  is  in  the  form  of  a  conic  frustrum  ;  the  bot- 
tom of  it  is  2  inches  in  diameter,  the  top  3,  and  the  depth 
6.7365  ;  this  crucible  is  filled  with  melted  metal,  of  which  it 
is  required  to  make  a  sphere ;  what  is  the  diameter  of  the 
mold  ?  Ans.  4  inches. 

45.  Suppose  a  cistern  has  two  pipes,  and  that  one  can  fill 
it  in  8^  hours,  the  other  in  4| ;  in  what  time  can  both  fill  it 
together'?  Ans.  3  hours  2  min.  49.5  sec. 

46.  A  certain  cistern  has  three  pipes ;  the  first  will  empty 
it  in  20  minutes,  the  second  in  40  minutes,  and  the  third  in 
75  minutes ;  in  what  time  would  they  all  empty  it  ? 

Ans.  11  min.  19  sec.  15  thirds. 

*  A  cubic  foot  of  sea  water  weighs  64  pounds. 

t  The  weight  of  a  cubic  foot  of  marble  is  assumed  to  be  185  pounds. 

%  The  weight  of  a  cubic  foot  of  fresh  water  is  62.5  pounds. 


MENSURATION    OF  ^SOLIDS.  217 

47.  A  reservoir  of  water  has  two  supply  cocks ;  the  first 
will  fill  it  in  40  minutes,  and  the  second  in  50 ;  it  has  also  a 
discharging  cock,  by  which  it  may  be  emptied,  when  full,  in 
25  minutes.  Now  if  all  the  cocks  are  opened  at  once,  and 
the  water  runs  uniformly,  how  long  before  the  cistern  will  be 
filled'?  Ans.  3  hours  20  minutes. 

Operation.  If  it  will  fill  once  in  40  minutes  by  one  cock,  and  also  in 
50  minutes  by  the  other  cock,  it  will  fill  2.25  times  in  50  minutes  by 
both  cocks. 

Then,  2.25  times :  50  minutes'.  '.  1  time :  22.222  minutes  — the  time  in  which 
it  will  Jill  by  both  cocks. 

Now,  asdt  will  empty  in  25  minutes,  the  time  gained  in  filling  over 
emptying  will  be  as  25  to  22.222,  or  9  to  8. 

Consequently,  it  will  fill  9  times  while  it  empties  8  times ;  and  as 
the  time  of  filling  is  22.222  minutes,  22.222x9  =  199.998  minutes  =  the 
time  required. 

48.  A  cistern  containing  60  gallons  of  water  has  three  un- 
equal cocks  for  discharging  it ;  the  largest  will  empty  it  in  one 
hour,  the  second  in  two  hours,  and  the  third  in  three ;  in  what 
time  will  the  cistern  be  emptied  if  they  all  run  together  % 

Ans.  32  min.  43  sec.  26  thirds. 
Operation. 
60  galls,  by  smallest  cock  in  3  hours  =-%°-= 20  galls,  in  1  hour. 
60     "      "second    cock  in  2  hours=4|>=30       "       1     " 
60     "      "  largest    cock  in  1  hour  =-S,G=60       "        1     " 
Then,  20+30+60=110  gallons  in  1  hour. 

Hence,  110  gallons:!  hour'.: 60  gallons :  .5454  hours =32  minutes,  43 
seconds,  and  26  thirds—the  result  required. 

49.  A  reservoir  has  three  pipes;  the  first  can  fill  it  in  12 
days,  the  second  in  11  days,  and  the  third  can  empty  it  in  14 
days  ;  in  what  time  will  it  be  filled  if  they  are  all  running  to- 
gether? Ans.  9  days  17  hours  24  min. 

50.  If  a  pipe  1J  inches  in  diameter  will  fill  a  cistern  in  50 
minutes,  how  long  would  it  require  a  pipe  that  is  2  inches  in 
diameter  to  fill  the  same  cistern  ?  Ans.  28  min.  7.5  sec. 

51.  If  a  pipe  6  inches  in  diameter  will  draw  off  a  certain 
quantity  of  water  in  4  hours,  in  what  time  would  it  take  3 
pipes  of  four  inches  in  diameter  to  draw  off  twice  the  quan- 
tity %  Ans.  6  hours. 

K 


218  MENSURATION    OF    SOLIDS. 

52.  A  water  tub  contains  147  gallons;  the  supply  pipe 
gives  14  gallons  in  9  minutes ;  the  tap  discharges  40  gallons 
in  31  minutes ;  now,  supposing  the  tap,  the  tub  being  empty, 
to  be  carelessly  left  open,  and  the  water  to  be  turned  on  at  2 
o'clock  in  the  morning  ;  a  servant  at  5,  finding  the  water  run- 
ning, shuts  the  tap.  Required  the  time  in  which  the  tub  will 
be  filled  after  this  discovery. 

Ans.  6  hours  3  min.  48.7  sec. 

53.  If  the  diameter  of  the  earth  is  7930  miles,  and  that 
of  the  moon  2160,  required  the  ratio  of  their  surfaces  and 
their  solidities,  assuming  them  to  be  spheres. 

Note. — The  surfaces  of  all  similar  solids  are  to  each  other  as  the 
squares  of  their  like  dimensions,  such  as  diameters,  circumferences, 
linear  sides,  etc.,  etc. ;  and  their  solidities  are  as  the  cubes  of  those 
dimensions. 

Operation.  Hence,  the  surface  of  the  moon :  the  surface  of  the  earth 

Also,  the  solidity  of  the  moon :  solidity  of  the  earth ::  21603 :  79303, 
21fi03  1 

54.  A  sugar-loaf  is  to  be  divided  equally  among  three  per- 
sons by  sections  parallel  to  the  base ;  it  is  required  to  find  the 
height  of  each  person's  share,  assuming  the  loaf  to  be  a  cone 
the  height  of  which  is  20  inches. 

203 
Operation.  By  similar  cones,  f3: 1::203:  — =2666. 667 =</*e  cube  of 

o 

the  height  of  the  upper  section;  hence,  -^2666.667  =  13.867,  the  upper  part. 
ov  203 
Also,  3:2:  :203: — —-=5333.333,  and  ^5333.333-13.867=3.604, 


the  middle  part;  consequently,  the  lower  part  will  be  20—13.867+3.604 
=  2.529  inches. 


*  The  ratio  of  one  quantity  to  another  may  be  obtained  by  dividing 

the  antecedent  by  the  consequent. 

t  This  proportion,  as  well  as  all  others  of  the  kind,  may  be  ex- 

20 
pressed  thus :  v'  3 :  y/ 1 : :  20 :  -^-^  =  the  height  of  the  upper  section ;  and 

V  o 

in  some  instances  this  ii  the  most  convenient  method. 


MENSURATION    OF   SOLIDS.  219 

55.  A  ship  has  a  leak  by  which  she  would  fill  and  sink  in 
15  hours,  but  by  means  of  her  pumps  she  can  be  pumped  out, 
if  full,  in  16  hours ;  now,  if  the  pumps  are  worked  from  the 
time  the  leak  begins,  how  long  before  the  ship  will  sink  ? 

Ans.  240  hours.  ■* 

Operation.  She  will  fill  -^  in  an  hour ;  then,  if  -fa  is  pumped  out, 
the  water  gains  -j^— ■£$= ^io  °f tne  sniP  Per  hour. 

56.  Three  men  bought  a  piece  of  tapering  timber,  which 
was  the  frustrum  of  a  square  pyramid ;  one  side  of  the  base 
was  3  feet,  one  side  of  the  top  1  footj  and  the  length  18  feet ; 
what  is  the  length  of  each  man's  piece,  assuming  they  are  to 
divide  equally  ? 

Operation.  By  similar  triangles,  3  —  1  :  18:  :1  :  9  =  the  length  of  the 
piece  cut  off  from  the  end,  by  which  the  piece  is  deficient  of  being  a 
pyramid. 

Then,  by  rules,  pages  168, 169,  the  contents  of  the  piece  of  timber  (a 
frustrum  of  a  pyramid)  and  the  piece  cut  off  (a  pyramid)  are  78  and  3 
cubic  feet.  Also,  78-^-3  = 26  —  the  contents  of  each  person' s  share,  which, 
a(Jded  to  the  contents  of  the  piece  cut  off=26  +  3=29=*fo  contents  of 
the  first  division  (including  the  piece  cut  off),  and  26  X  2 +3 =55  =  the  con- 
tents of  the  second  division  (including  the  piece  cut  off). 

Now,  by  similar  pyramids,  3:29::93  (length  of  piece  cut  off):70<!t7, 
and  V  7047  =  19.172;  hence,  19.172 -9  =  10.172  =  the  length  of  the  first 
division  of  the  frustrum. 

Again,  29  :  55: :  19.1723  (length  of  first  division) :  13365,  and  v7  13365 
=23.731;  hence,  23.731  -19. 172  =4.559 =the  length  of  the  second  di- 
vision of  the  frustrum. 

Hence, 

Length  of  timber... 18.000 

"  first  division 10.172 

"  second  division 4.559  14.731 

"  third  division 3.269 

57.  Assuming  the  earth  to  be  a  sphere,  and  a  quarter  of 
its  radius  1000  miles,  what  is  the  area  of  its  surface,  its  vol- 
ume, and  weight,  the  mean  density  of  it  being  353.75  pounds 
per  cubic  foot  ? 

i       $  Surface,  201062400  square  miles. 
US'  {.Volume,  268083200000  cubic  miles. 


220  MENSURATION    OF    SOLIDS. 

58.  The  sides  of  the  base  of  an  irregular  tetrahedron  are 
21,  20,  and  13  feet,  and  its  height  9  ;  what  is  its  volume? 

By  rules,  pages  56  and  60,  the  area  of  the  base  of  the  figure  is  12 
feet. 

Ans.  756  cubic  feet 

59.  A  regular  tetrahedron  contains  1.8414  cubic  feet;  re- 
quired its  side  and  surface. 

Ans.  Side,  30  inches ;  surface,  1558.845  square  inches. 

60.  Required  the  volume  of  the  frustrum  of  a  triangular 
pyramid,  the  base  of  which  has  a  side  of  9  inches,  its  vertex 
4,  and  its  lateral  edge  5. 

Operation.  It  is  required  to  ascertain  the  perpendicular  height  or 
length  of  the  frustrum. 

1.  92-(9-^2)2  =  60.75,  and  V  60. 7 5 =7. 7942  =  perpendicular  of  trian- 
gle of  base. 

2.  By  rule,  p.  62,  to  ascertain  centre  of  base  (an  equilateral  triangle), 
9  X. 5773  —  5.1957= radius  of  circumscribing  circle  of  base,  or  centre  of 
base. 

3.  The  base  and  vertex  of  the  frustrum  have  sides  of  9  and  4,  and  the 
slant  height  is  5  ;  hence,  9co4:  5 '.'A:  4:=  slant  height  of  end  of  pyramid 
if  continued ;  and  5  +  4=9  =height  of  whole  pyramid.  • 

4.  If  5.196  =radius  of  base  of  pyramid,  and  9=its  slant  height,  then, 
V (5A962-9*)=7M8=height  of  ichole pyramid;  and 9: 5:: 7.348  : 4.082 
= height  of  frustrum  of  pyramid. 

Then,  by  rule,  p.  169,  92+42+9  x4  X  .433 X 4.082-r-3=78.3595  =  ?e- 
sult  required. 

6 1 .  The  sides  of  the  base  of  an  irregular  tetrahedron  are 
12,  15,  and  17  inches,  and  its  height  9  ;  required  its  volume. 

Ans.  263.248  cubic  inches. 

62.  How  large  a  cube  may  be  inscribed  in  a  sphere  40 
inches  in  diameter?  Ans.  23.094  inches. 

63.  How  many  cubic  inches  are  contained  in  a  cube  that 
may  be  inscribed  in  a  sphere  20  inches  in  diameter  ? 

Ans.  1480.2936  cubic  inches. 
By  rule,  page  61. 

64.  If  a  stone  is  put  into  a  vessel  of  14  cubic  feet  in  capac- 
ity, and  it  then  requires  but  2.5  quarts  of  water  to  fill  it,  what 
is  the  volume  of  the  stone? 

Ans.  13  feet  1560  inches. 


MENSURATION   OF    SOLIDS.  221 

65.  A  cone,  the  diameter  of  which  is  12  inches  and  altitude 
10,  being  put  into  a  vessel  filled  with  rain  water,  with  its  base 
upward,  was  depressed  to  a  point  where  the  area  of  its  sec- 
tion, parallel  to  the  base,  was  eighty  inches ;  required  the 
weight  of  the  cone. 

As  12  :  10::  ■/ (80-4-. 7854)  :  8.41  inches=depth  of  the  cone  in  water; 
then,  80x8.41x^=224.266,  cubic  inches,  the  volume  of  the  part  im- 
mersed.    Consequently,  as  1728  :  224.266 : :  1000*  :  129.784  ounces. 

66.  Into  a  vessel  filled  with  rain  water,  suppose  there  be  put 
a  cone  of  dry  wood,  having  a  volume  of  one  cubic  foot,  with 
the  lesser  end  downward,  and  its  axis  perpendicular  to  the 
surface  of  the  water,  and  if  a  plane  passing  through  the  cen- 
tre of  gravity  of  the  piece,  parallel  to  its  base,  coinciding  with 
the  water's  surface,  is  found  to  rest  in  equilibrium,  required 
the  quantity  of  water  that  will  run  over  from  the  vessel,  and 
the  specific  gravity  of  the  cone. 

Operation. — The  centre  of  gravity  of  a  cone  is  distant  from  the  ver- 
tex f  of  its  axis ;  hence,  43 :  33  .* :  1728 :  729  =cubic  inches  immersed  in  the 
water. 

^Consequently,  the  quantity  of  water  run  over  will  be  729  cubic  inches, 
and  the  specific  gravity  of  the  water  will  be  to  that  of  the  cone  as  729 
to  1728,  or  as  27  to  64,  Ans. 

67.  A  right  cone  cost  $1363,  at  $120  per  cubic  foot,  the 
diameter  of  its  base  being  to  its  altitude  as  5  to  8.  It  is  re- 
quired to  have  its  convex  surface  divided  in  the  same  ratio  by 
a  plane  parallel  to  the  base,  the  upper  part  to  be  the  greater ; 
what  is  the  slant  height  of  each  part  ? 

1363—120  =  1  1.358= the  volume  of  the  cone  in  feet;  and  53X.7854x 
•§•=52.36 =volume  of  a  cone  similar  to  it,  the  altitude  of  which  is  8. 

Again,  the  surface  of  similar  solids  being  as  the  squares  of  their  like 
dimensions,  \/(5  +8) :  V8 : :  \/(2.52  +  82)  the  side  of  the  said  similar  cone 
:  V&ffi  =  the  slant  height  of  the  upper  part  of  this  cone  when  its  surface  is 
divided  in  the  ratio  proposed. 

Consequently,  V  70^-V  ^-=V  ^-V  ^-=the  slant  height  of  the 
under  part  of  it. 

Then,  as  similar  solids  are  to  each  other  as  the  cubes  of  their  like  di- 
mensions,  ^52.36:  / 11.358: :V^-:S.9506  =  the  length  of  the  slant 
height  of  the  upper  part. 

*  Weight  of  a  cubic  foot  of  fresh  water  in  ounces. 


222       -  MENSURATION    OF   SOLIDS. 

Again,  ^52.36  :  f  11.358 : :  V&&-V ^f* :  1.0855  =  *Ae  length  of  the 
slant  height  of  the  lower  part. 

68.  An  elliptic  inclosure  has  diameters  of  £30  and  612 
links  within  its  wall,  which  is  14  inches  thick ;  required  the 
area  it  incloses  and  covers. 

a       jit  incloses  4  acres  and  6  poles,  and 
'  \        covers  1760.5  square  feet. 

69.  A  block  of  marble,  in  the  form  of  a  square  pyramid, 
weighs  18  tons,  the  perpendicular  height  being  twice  the  di- 
agonal of  the  base ;  required  its  dimensions  and  volume  in 
cubic  feet. 

A  cubic  inch  of  the  marble  is  assumed  to  weigh  1.6  ounces  avoir- 
dupois. 

A        (Side,  4.35325  feet. 
nS'  t  233.334  cubic  feet. 

70.  A  pail  containing  2.1215  cubic  feet  is  12  inches  in 
depth ;  what  are  its  top  and  bottom  diameters,  they  being  in 
the  proportion  of  5  to  3  ? 

Ans.  14.64  and  24.4  inches. 

71.  If  a  sphere,  the  diameter  of  which  is  4  inches,  is  de- 
pressed in  a  conical  glass  full  of  water,  the  diameter  of  which 
is  5  inches  and  altitude  6,  it  is  required  to  know  the  quantity 
of  water  which  will  run  over?  Ans.  26.272  inches. 

By  Construction,  Fig.  132. 

Draw  a  section  of  the  glass,  as  A  B  C,  and  of  the  sphere,  d  ef. 

Fig.  132.  e 


MENSURATION    OF    SOLIDS. 


223 


Then  will  AB=5;  Ci  =  6;  Bt'  =  54-2  =  2.5;  A  C  and  B  C  = 
V(C  i2+B  i2)=6.5;  and  o/=4-j-2  (half  diameter  of  sphere) =2. 

The  triangles  B»C  and  o  f  C  are  similar,  for  they  have  the  com- 
mon angle  C,  and  the  right  angles  i  andy;  hence,  their  remaining  an- 
gles are  equal. 

Therefore,  B  i :  B  Cl'.of-.o  C ;  that  is,  as  2.5  : 6.5 : : 2  :  5.2  =  the  depth 
of  the  glass  from  where  the  centre  of  the  sphere  rests. 

Consequently,  5.2  — 6  =  .8,  and  2  +  .8 =2. 8  inches  of  the  sphere  immersed, 
and  4  —  2.8  =  1.2  inches  of  it  above  the  glass. 

Hence,  the  volume  of  the  segment  immersed  (hy  rule  2,  p.  177)  de- 
ducted from  that  of  the  sphere =26.2722  cubic  i?iches  =  the  result  re- 
quired. 

72.  If  a  sphere  is  depressed  in  a  conical  glass  full  of  water, 
the  diameter  of  which  is  5  inches  and  altitude  6,  so  that  its 
upper  edge  is  in  a  line  with  the  rim  of  the  glass,  while  its  side 
rests  upon  the  side  of  the  glass ;  required  the  quantity  of  wa- 
ter that  will  overflow  from  the  glass. 

By  Construction,  Fig.  133. 

Draw  a  section  of  the  glass,  as  A  B  C,  and  of  the  sphere,  as  d  ef. 


Fig.  133. 


Then  will  C  c=6,  B  C=f =2.5,  and,  by  preceding  example,  B  C 
=6.5. 

Hence,  o  e  and  of  are  equal,  and  the  line  B  o,  bisecting  B  e  o  f,  is 
a  hypothenuse  common  to  both  triangles,  B  e  o,  Hfo. 

Consequently,  B  c  and  B  f  are  equal  to  2.5,  and  B  C— ~B  f=C  f= 
6.5-2.5  =  4. 

/.  6  (C  e) :  2.5  (e  B) : :  4  (C/) :  1.667  (/o),  and  1.667  X  2=3.334  inches, 
the  diameter  of  the  sj>here,  the  volume  of  which  (by  rule,  p.  176)  =  19.387 
inches = the  quantity  of  water  that  will  over/low  from  the  glass. 


224  MENSURATION   OF    SOLIDS. 

73.  If  a  sphere,  the  diameter  of  which  is  4  inches,  is  de- 
pressed in  a  conical  glass  -L  full  of  water,  the  diameter  of 
which  is  5  inches  and  altitude  6,  it  is  required  how  much  of 
the  vertical  axis  of  the  sphere  is  immersed  in  water. 

By  Construction,  Fig.  134. 

Draw  a  section  of  the  glass,  as  A  B  C,  and  of  the  sphere,  d  ef 

Fig.  134.  __i_ 


Let  m  n  be  the  original  level  of  the  water,  and  n  r  the  level  when  the 
sphere  is  immersed. 

Then  will  the  cone  n  C  r=cone  m  C  n+the  volume  of  the  segment  of 
the  sphere  d  sf=^  of  cone  A  B  C-\-the  volume  of  the  segment  d  sf. 

A  C  =  VCi2(6)+Ai2(5+2)=6.5  =  length  of  slant  side. 

As  A  i  (2.5) :  A  C  (6.5) : :  of  radius  of  sphere  (2) :  C  o  {o.2)=dis- 
tancefrom  the  centre  of  sphere  at  rest  and  the  bottom  of  the  glass. 

C  s=C  o-o  s=5.2-2=3.2. 

Contents  of  cone  (by  rule,  p.  166)=39.27,  •£  of  which =7.854. 

Put  x=s  u—the  immersed  part  of  the  axis  of  the  sphere,  and  C  w=C  s 
+s  u=S.2+x. 

Then,  as  similar  solids  are  to  each  other  as  the  cubes  of  their  like  di- 
mensions, 

63:  (3.2 +:r)3::  39.27:  cone  nCr;     .'.cone  n  C  r^^^x  39.27. 

216 

Segment  dsf  (by  rule  2,  p.  177)=(4x3t-2a;)Xa:2X.5236. 

(3  2+a:)3 
Since  Cone  n  C  r=cone  m  C  n+segment  d  sf,  .'.— X 39.27= 

7.854+(4  x  3-2z)  x  x2  x  .5236. 

And  25  X  (3.2 +x)3=5  X  216  +  4  X  62  x  (6  -x)  X  x2. 

Cube  3.2+x  and  25  (3.23+3x3.22x+3x3.2xs+x3)=5x63+4x6» 
(6-x)Xx2. 


MENSURATION   OF    SOLIDS.  225 

Actually  multiplying  the  terms  in  the  first  member  by  25,* 

163 

— +3  •  162x+3  •  5  •  16x2+25x3=5  •  63+4  •  62  (6-x)x'j 
5 

But  4  •  62  (6-x)x*=4:  •  63o:2-4  ■  62*3. 

lfi3 
/.—-+3  •  162o;+3  •  5  ■  1 6a;5 -f  25a:3 =5  '  63+4  •  63x2-4  •  62x3. 
5 

Multiplying  by  5  throughout, 

163+3  •  5  •  162x+3  •  52  •  16a;2 +5V =52  63+4  ■  5  •  63x2-4  •  5  ■  62x3. 

By  transposition, 

(53-4  •  5  •  62)*3+(3  ;  52  •  1G-4  •  5  •  63)x3  +  3  •  5  •  162o;=52  •  63-163. 

But,  S=  +  4  ■  5  •  G'=^p=169  =  f  ±jjj  X5  =  13»x5. 

3-5-16-4-5-6»=-3120=-2^,^,^  =  16=3-5-13-16 

6  o        lo 

=3120. 
52  •  63-163  =  1304=-^=163=8  ■  163. 

3.5.162=8-^,^,^,  16=3-5. 162=3840. 
3  5        16 

.'.5  •  132o;3-3  •  5  •  13  •  16x3+3  -5  •  163ar=8  ■  163. 

.-.5  (132*3-3  •  13  •  16x2+3  •  162x)=8  ■  163. 

Or,  132o:3-3  ■  13  •  16x2+3  •  16'a;  =  — - — . 

5 

Multiplying  both  members  by  13, 

iSV-8  •  132  •  16x2+3  •  13  •  162*=8'13/163=^=3390. 


Subtract  163  from  both  numbers 
lS3x*  -  3  •  132  '  16a:2  +  i 
8-13-163-5-163     -3528 


133x3  -  3  •  132  •  16a:2  +  3  ■  13  •  162a:  -  163  = 16»  = 


705-6. 


5  5 

The  first  member  is  now  a  perfect  cube,  the  root  of  which  is 
13a; - 16  =  -^  -705.6=8.90265. 
13x= 16  -  8.90265 =7.09735. 

*  =         —  =  .54595  inch. 

13 

Proof.  C  s=3.2/and  s  w=.54595. 
Hence,  C  «=3.2  +  .54595=3.74595. 

*  By  Professor  G.  B.  Docharty,  New  York. 

f  See  Explanation  of  Characters,  page  10,  for  use  of  a  period  be- 
tween two  factors. 

K2 


226  MENSURATION    OF    SOLIDS. 

Then,  63 :  39.27 1 : 3.745953 : 9.5563 =volume  of  water  in  cone,  nCr,  from 
which  is  to  be  deducted  the  volume  of  segment  d  sf  and  the  remainder 
should  be  equal  to  \  of  39.27  =  7.854. 

Thus,  volume  of  segment  (by  rule  2,  p.  177)  =  1.7023,  and  9.5563 
— 1.7023  =  7.85 4= result  required. 

74.  A  lady  having  three  daughters  had  a  farm  of  450.758 
acres,  in  a  circular  form,  with  her  dwelling-house  in  the  cen- 
tre. Being  desirous  of  having  her  daughters  near  her,  she 
gave  to  them  three  equal  parcels  of  land  as  large  as  could  be 
made  in  three  equal  circles  within  the  periphery  of  her  farm, 
one  to  each,  with  a  dwelling-house  in  the  centre  of  each ;  that 
is,  there  were  to  be  three  equal  circles  as  large  as  could  be 
drawn  within  the  periphery  of  the  farm ;  required  the  diam- 
eters of  the  farm  and  of  the  three  parcels. 


V450.758  x43560-r-.7854=5000/ee*=dww»efer  of  farm. 
By  Construction,  Fig.  135. 


Draw  the  given  circle,  with  o  as  its  centre,  and  divide  its  periphery 
into  three  equal  parts,  as  at  A  B  C ;  connect  A  o,  B  o,  and  C  o,  and 
assume  d  ef&s  the  centres  of  the  required  circles. 

As  the  three  circles  required  touch  one  another  and  the  given  circle, 
the  points,  as  A,  B,  and  C,  the  centres,  d  ef,  of  the  required  circles, 
and  o,  are  necessarily  in  right  lines.     Connect  d,  ^  &ndf. 

Then,  as  d  ejfand  e  of  axe  isosceles  triangles,  the  angle  d  and  the 
base  e/are  bisected  at  right  angles  in  i  by  the  line  d  i,  and  e  o,  in  like 
manner,  bisects  the  angle  e. 

The  triangles,  e  di,  e  o  i,  are  equiangular: 

Hence,  ed:ei=($ed)::e  o:oi=($e  o);   :.e  o=2o  i. 


MENSURATION    OF    SOLIDS.  227 

Put  B  e=x=(e  i),  R=B  o=2500=radius  of  given  circle  =  (  j ; 


„  i  .    R— * 

then,  e  o=K—x.      :.^eo=oi= — - — 


eoa=et9+o»8,or(R-a;)a=ara+ 
Or,  R2-2Rx+x2=a;2  + 


2 

(R-*)2 


4 
R2-2R*+a:2 


4 

Or,  4R2-8Ra;=R2-2Ra:+a;2. 

Transposing  the  formulae, 

x2  +  6Rar=3R2:'x2  +  6Rx+9R2=3R2+9R2=12R?. 

x+3R=±V12R2:  x=±-/l2R1'-3R 

x  s=  ±Ra/  1 2  —  3R :  x  =  ±R( \/ 12 — 3) = radius  of  required  circles. 

.'.Radius  being=-/12— 3=3.4641— 3  =  .4641,  wfo'c^  is  a  constant 
multiplier  for  all  like  problems. 

Consequently,  2500  x  .4641  =  1160.25=A  d,  B  e,  and  Cf=product  of 
radius  of  circle  and  multiplier ■=radii  of  each  of  the  circles  required. 

75.  The  weight  of  a  quantity  of  silt  in  30  cubic  inches  of 
salt  water  is  4.21  grains,  assuming  the  weights  of  silt  and 
salt  water  to  be  respectively  125  and  64  pounds  per  cubic 
foot ;  what  is  the  volume  of  the  silt  compared  to  that  of  the 
water?  Am.  1.  to  3608.307. 


228  CONIC   SECTIONS. 


CONIC  SECTIONS. 

Definition.     Plane  figures  generated  by  the  cutting  of  a  cone. 

A  Cone  is  a  figure  described  by  the  revolution  of  a  right- 
angled  triangle  about  one  of  its  legs. 

c 


The  axis  (of  a  cone)  is  the  line  about  which  the  triangle  re- 
volves, as  C  o. 

The  base  is  the  circle  which  is  described  by  the  revolving 
base  of  the  triangle,  as  B  o. 

Notes. — If  a  cone  is  cut  by  a  plane  through  the  vertex  and  base,  the 
section  will  be  a  triangle,  as  A  C  B. 

If  a  cone  is  cut  by  a  plane  parallel  to  its  base,  the  section  will  be  a 
circle. 

An  Ellipse  is  a  figure  generated  by  an  oblique  plane  cutting 
a  cone,  as  a  b  c  d.  v 


The  transverse  axis  or  diameter  (of  an  ellipse)  is  the  longest 
right  line  that  can  be  drawn  in  it,  as  a  b. 

The  conjugate  axis  or  diameter  is  a  line  drawn  through  the 
centre  of  the  ellipse  perpendicular  to  the  transverse  axis,  as  c  d. 


CONIC    SECTIONS.  229 

A  Parabola  is  a  figure  generated  by  a  plane  cutting  a  cone 
parallel  to  its  side,  as  a  b  c. 


The  axis  (of  a  parabola)  is  a  right  line  drawn  from  the  ver- 
tex to  the  middle  of  the  base,  as  b  o. 

Note. — A  parabola  has  no  conjugate  diameter. 

An  Hyperbola  is  a  figure  generated  by  a  plane  cutting  a 
cone  at  any  angle  with  the  base  greater  than  that  of  the  side 
©f  the  cone,  as  a  b  c. 


ft-- 


«       jg 


The  transverse  axis  or  diameter,  o  b  (of  an  hyperbola),  is  that 
part  of  the  axis  e  b,  which,  if  continued,  as  at  o,  would  join 
an  opposite  cone,  ofr. 

The  conjugate  axis  or  diameter  is  a  right  line  drawn  through 
the  centre,  g,  of  the  transverse  axis,  and  perpendicular  to  it. 

The  straight  line  through  the  foci  is  the  indefinite  trans- 
verse axis ;  that  part  of  it  between  the  vertices  of  the  curves, 
as  o  bf  is  the  definite  transverse  axis.  Its  middle  point,  g,  is 
the  centre  of  the  curve. 


230 


CONIC    SECTIONS. 


The  eccentricity  of  an  hyperbola  is  the  ratio  obtained  by  di- 
viding the  distance  from  the  centre  to  either  focus  by  the  semi- 
transverse  axis. 

The  asymjrtotes  of  an  hyperbola  are  two  right  lines  to  which 
the  curve  continually  approaches,  touches  at  an  infinite  dis- 
tance, but  does  not  pass ;  they  are  prolongations  of  the  diag- 
onals of  the  rectangle  constructed  on  the  extremes  of  the  axes. 

Two  hyperbolas  are  conjugate  when  the  transverse  axis  of 
the  one  is  the  conjugate  of  the  other,  and  contrariwise. 

An  Ordinate  is  a  right  line  from  any  point  of  a  curve  to 
either  of  the  diameters,  as  a  e  and  do;  ab  and  dfare  double 
ordinates. 


c  d 

e  i 

—\ 

i     o 


An  abscissa  is  that  part  of  the  diameter  which  is  contained 
between  the  vertex  and  an  ordinate,  as  c  e,  g  o. 

The  parameter  of  any  diameter  is  equal  to  four  times  the 
distance  from  the  focus  to  the  vertex  of  the  curve ;  the  param- 
eter of  the  axis  is  the  least  possible,  and  is  termed  the  param- 
eter of  the  curve. 

The  parameter  of  the  curve  of  a  conic  section  is  equal  to 
the  chord  of  the  curve  drawn  through  the  focus  perpendicular 
to  the  axis. 

The  parameter  of  the  transverse  axis  is  the  least,  and  is 
termed  the  parameter  of  the  curve. 

The  parameter  of  a  conic  section  and  the  foci  are  sufficient 
elements  for  the  construction  of  the  curve. 

In  the  Parabola  the  parameter  of  any  diameter  is  a  third  proportional 
to  the  abscissa  and  ordinate  of  any  point  of  the  curve,  the  abscissa  and 
ordinate  being  referred  to  that  diameter  and  the  tangent  at  its  vertex. 


CONIC   SECTIONS. 


231 


In  the  Ellipse  and  Hyperbola,  the  parameter  of  any  diameter  is  a  third 
proportional  to  the  diameter  and  its  conjugate. 

Note. — To  determine  the  Parameter  of  an  Ellipse  or  Hyperbola. 

Kule.  Divide  the  product  of  the  conjugate  diameter,  multiplied  by  it- 
self, by  the  transverse,  and  the  quotient  is  equal  to  the  parameter. 

In  the  annexed  figures  of  an  Ellipse  and  Hyperbola,  the  transverse  and 
conjugate  diameters,  ab,  c  d,  are  each  30  and  20. 


TJien,  30 :  20 : :  20 :  13.333  ^parameter. 


Hence,  the  parameter  of  the  curve =ef  a  double  ordinate  passing 
through  the  focus  s. 

In  a  Parabola.     The  abscissa  a  b,  and  ordinate  c  b,  are  also  equal 
to  30  and  20. 

a 


c  b 

Hence,  the  parameter  of  the  curve =e/. 


232 


CONIC    SECTIONS. 


A  Focus  is  a  point  on  the  principal  axis  where  the  double 
ordinate  to  the  axis,  through  the  point,  is  equal  to  the  pa- 
rameter, ase/  in  the  preceding  figures. 

It  may  be  determined  arithmetically  thus :  Divide  the  square  of  the 
ordinate  by  four  times  the  abscissa,  and  the  quotient  will  give  the  focal 
distances  a  s  and  s  in  the  preceding  figures. 

The  Directrix  of  a  conic  section  is  a  straight  line,  such  that 
the  ratio  obtained  by  dividing  the  distance  from  any  point  of 
the  curve  to  it  by  the  distance  from  the  same  point  to  the  fo- 
cus shall  be  constant. 

It  is  always  perpendicular  to  the  principal  axis ;  and  if  the 
curve  is  given,  it  is  constructed  as  follows : 


Let  ABC  represent  the  curve  or  curves,  e  f  their  axis, 
and  s  the  focus. 

Through  s  draw  s  n  or  n'  perpendicular  to  the  axis  till  it 
meets  the  curve  in  n  or  n' ;  at  n,  n'  draw  the  tangents  weor 
n'  e%  cutting  the  axis  at  e  and  e' ;  through  e,  ef  draw  g  h  and 
g'  h'  perpendicular  to  the  axis,  and  they  will  be  the  directrices 
of  two  conic  sections. 

If  d  s,  drawn  from  any  point,  as  d,  >  (is  less  than)  d  d,  the 
curve  is  an  ellipse ;  if  equal  to  each  other,  it  is  the  curve  of  a 
parabola ;  and  <  (if  greater),  as  d/  s,  d/  d/,  it  is  the  curve  of 
an  hyperbola. 

Ellipsoid,  Paraboloid,  and  Hyperboloid  of  Revolution.  Fig- 
ures generated  by  the  revolution  of  an  ellipse,  parabola,  etc., 
around  their  axes.     (See  p.  124  and  202.) 


CONIC    SECTIONS. 


233 


A  Conoid  is  a  warped  surface  generated  by  a  right  line  be- 
ing moved  in  such  a  manner  that  it  will  touch  a  straight  line 
and  curve,  and  continue  parallel  to  a  given  plane.  The  straight 
line  and  curve  are  called  directrices,  the  plane  a  plane  direc- 
trix, and  the  moving  line  the  generatrix. 


Z\ 


j7 


Thus,  let  a  b  c  be  a  circle  in  a  horizontal  plane,  and  d  d' 
the  projection  of  a  right  line  perpendicular  to  a  vertical  plane, 
d  e  ;  if  right  lines,  d  a,  r  s,  r'  b,  r"  s,  and  df  c,  be  moved  so 
as  to  touch  the  circle  and  right  line  d  d/,  and  be  constantly- 
parallel  to  the  plane  r  b,  it  will  generate  the  conoid  d  b  s  n. 

Note. — All  the  figures  which  can  possibly  be  formed  by  the  cutting 
of  a  cone  are  mentioned  in  these  definitions,  and  are  the  five  following, 
viz.,  a  triangle,  a  circle,  an  ellipse,  a  parabola,  and  an  hyperbola;  but  the 
last  three  only  are  termed  the  conic  sections. 


234 


CONIC    SECTIONS. 


ELLIPSE. 

To  describe  an  Ellipse. 
The  Transverse  and  Conjugate  Diameters  being  given,  Fig.  1. 

Rule  1. — Draw  the  transverse  and  conjugate  diameters, 
A  B,  C  D,  bisecting  each  other  perpendicularly  in  o. 

Make  A  e  equal  to  D  C ;  divide  e  B  into  three  equal  parts ; 
set  off  two  of  those  parts  from  o  to  e  and  from  o  to  c ;  then 
with  the  distance  c  e  make  the  two  equilateral  triangles  c  b  e 
and  q  d  e. 

These  angles  are  the  centres,  and  the  sides  being  contin- 
ued are  the  lines  of  direction  for  the  several  arcs  of  the  ellipse 
A  C  B  D. 

Fig.  1. 

J 


Note. — Mechanics  are  oftentimes  required  to  work  an  architrave, 
etc.,  about  windows,  of  this  form ;  they  may,  by  the  help  of  the  four 
centres  c,  d,  e,  b,  and  the  lines  of  direction  h  d,  b  f  d  g,b  i,  describe 
another  ellipse  around  the  former,  and  at  any  distance  required. 

Rule  2. — Draw  the  line  C  D  equal  in  length  to  the  trans- 
verse diameter ;  also,  E  F  equal  in  length  to  the  conjugate  di- 
ameter, and  at  right  angles  with  C  D. 

Take  the  distance  C  o  or  o  D,  and  with  it,  from  the  points 
E  and  F,  intersect  the  diameter  C  D  at  h  and  f  which  points 
are  the  foci. 

Secure  a  string  at  h  and  f  of  such  a  length  that  it  may  just 
reach  to  E  or  F. 

Introduce  a  pencil,  and  bearing  upon  the  string,  carry  it 
around  the  centre  o,  and  it  will  describe  the  ellipse  required.* 

*  It  is  a  property  of  the  ellipse  that  the  sum  of  two  lines  drawn  from 
the  foci  to  meet  in  any  point  in  the  curve  is  equal  to  the  transverse  di- 


CONIC    SECTIONS. 


235 


Fig.  2. 


The  Transverse  Diameter  alone  being  given,  Fig.  3. 

Rule. — Let  A  B  be  the  given  length. 

Divide  it  into  three  equal  parts,  as  A  s  i  b.  Then,  Vith 
the  radius  A  s,  describe  A  /  o  i  n  c,  and  from  i  the  circle 
~B  dn  s  o  e ;  then  with  n  f  and  o  c  describe  f  e  and  c  d,  and 
the  required  ellipse  is  made. 

Fig.  3. 


When  any  three  of  the  four  following  Terms  of  an  Ellipse  are 
given,  viz.,  the  Transverse  and  Conjugate  Diameters,  an  Or- 
dinate, and  its  Abscissa,  to  find  the  remaining  Term. 

To  ascertain  the  Ordinate,  the  Transverse  and  Conjugate  Diam- 
eters and  the  Abscissa  being  given,  Fig.  4. 
Rule. — As  the  transverse  diameter  is  to  the  conjugate,  so 
is  the  square  root  of  the  product  of  the  two  abscissas  to  the 
ordinate  which  divides  them. 


Or,  -x  Vax(t— «0— °9  t  ^presenting  the  transverse  diam- 
eter, c  the  conjugate,  a'  the  less  abscissa,  and  o  the  ordinate. 

ameter,  and  from  this  the  correctness  of  the  above  construction  is  ev- 
ident. 


236 


Fig.  4. 


CONIC    SECTIONS. 
C 


Example. — The  transverse  diameter,  A  B,  of  an  ellipse, 
Fig.  4,  is  25  inches,  the  conjugate,  C  D,  is  16,  and  the  ab- 
scissa. A  i,  7 ;  what  is  the  length  of  the  ordinate  i  el 

25  —  7  =  18=  second  abscissa. 
Vl  x  18  =  11.225  = square  root  of  the  abscissce. 

Hence,  25  :  16::  11.225  :  7.184  inches,  the  length  of  the  ordinate  re- 
quired. 

Ex.  2.  The  transverse  diameter  or  axis  of  an  ellipse  is  100 
inches,  the  conjugate  60,  one  abscissa  20,  and  the  other  80 ; 
what  is  the  length  of  the  ordinate?  Ans.  24  inches. 


To  ascertain  the  Abscissa?,  the  Transverse  and  Conjugate  Diameters 
and  the  Ordinate  being  given,  Fig.  4. 
Rule. — As  the  conjugate  diameter  is  to  the  transverse,  so 
is  the  square  root  of  the  difference  of  the  squares  of  the  or- 
dinate and  semi-conjugate  to  the  distance  between  the  ordi- 
nate and  centre,  and  this  distance  being  added  to,  or  subtract- 
ed from  the  semi-transverse,  will  give  the  abscissas  required. 

~]  x   representing   the  dis- 

'    I      tame  obtained,  and  a 
'  the  greater  and  less 


-X=a>\      ab 


abscissa?. 


Example. — The  transverse  diameter,  A  B,  of  an  ellipse, 
Fig.  4,  is  25  inches,  the  conjugate,  C  D,  16,  and  the  ordinate 
t  e  7.184 ;  what  is  the  abscissa  «  B  ? 


Vj.  184*  —  82 =3.519943 —square  root  of  difference  of  squares  of  semi- 
conjugate  and  ordinate. 

Hence,  as  16 :  25 : :  3.52 : 5.5  ^distance  between  ordinate  and  centre. 


CONIC    SECTIONS.  237 

■        ok  _. 

Then,  y  =  12.5,  and  12.5  +5.5  =  18  =B  z,  | 

\  abscissas  required. 
-^  =  12.5,  and  12.5-5.5=  7=A  if,  f 

Ex.  2.  The  transverse  diameter,  A  B,  of  an  ellipse  is  50 
inches,  the  conjugate,  C  D,  32,  and  the  ordinate  i  e  14.368 ; 
what  are  the  lengths  of  the  abscissae  ? 

Ans.  11  and  39  inches. 


To  ascertain  the  Transverse  Diameter,  the  Conjugate,  Ordinate, 
and  Abscissa  being  given,  Fig.  4. 

Rule. — To  or  from  the  semi-conjugate,  according  as  the 
greater  or  less  abscissa  is  used,  adVjl  or  subtract  the  square  root 
of  the  difference  of  the  squares  of  the  ordinate  and  semi-con- 
jugate. 

Then,  as  this  sum  or  difference  is  to  the  abscissa,  so  is  the 
conjugate  to  the  transverse.  '* 

axe 


Or,c-f-2  + 

CH-2- 


W*ffl 


Example. — The  conjugate  diameter,  C  D,  of  an  ellipse, 
Fig.  4,  is  16  inches,  the  ordinate  i  e  is  7.184,  and  the  abscissas 
B  i,  i  A  are  18  and  7  ;  what  is  the  length  of  the  transverse 
diameter  ? 


^/7.1842—  (  —  J  =3.52  =square  root  of  difference  of  squares  of  ordi- 
nate and  semi-conjugate. 

y  +  3.52:18::16:25,l 

-g  >  =transverse  diameter  required. 

—  -3.52:    7::  16:  25,  J 

Ex.  2.  The  conjugate  diameter  of  an  ellipse  is  60  inches, 
the  ordinate  24,  and  the  abscissa  20 ;  what  is  the  length  of 
the  transverse  diameter?  Ans.  100  inches. 


238  CONIC    SECTIONS. 

To  ascertain  the  Conjugate  Diameter,  the  Transverse,  Ordinate, 
and  Abscissa  being  given,  Fig.  4. 

Rule. — As  the  square  root  of  the  product  of  the  abscissae 
is  to  the  ordinate,  so  is  the  transverse  diameter  to  the  con- 
jugate. 

Or,  oxt-i-  Vci  X  a' =c. 

Example. — The  transverse  diameter,  A  B,  of  an  ellipse, 
Fig.  4,  is  25  inches,  the  ordinate  i  e  7.184,  and  the  abscissae 
B  i  and  i  A  18  and  7  ;  what  is  the  length  of  the  conjugate  di- 
ameter % 


viSxl  =  11.225= square  root  of  product  of  abscissae. 
11.225  :  7.184:  :25  :  16=conjugate  diameter  required. 

Ex.  2.  The  transverse  diameter  of  an  ellipse  is  100  inches, 
the  ordinate  24,  and  the  abscissae  20  and  80 ;  what  is  the 
length  of  the  conjugate  diameter'?  Ans.  60  inches. 

To  ascertain  the  Circumference  of  an  Ellipse,  Fig.  4. 

Rule. — Multiply  the  square  root  of  half  the  sum  of  the 

squares  of  the  two  diameters  by  3.1416,  and  this  product  will 

give  the  circumference  nearly. 

/d2-\-d'2 
Or,  \/ —  x  3 . 1 4 1 6  =  circumference. 

Example. — The  transverse  and  conjugate  diameters,  A  B 
and  C  D,  of  an  ellipse,  Fig.  4,  are  24  and  20  inches ;  what  is 
its  circumference  I 

°4?  +  202 

- — =  488,  and  V  488 =22.09  =square  root  of  half  the  sum  of  the 

squares  of  the  diameters. 

Hence,  22.09  X3.1416=69.398=^e  above  rootx3.14:W=the  result  re- 
quired. 

Ex.  2.  The  diameters  of  an  ellipse  are  30  and  20  inches ; 
what  is  its  circumference?  Ans.  80.0951  inches. 


CONIC    SECTIONS. 


239 


To  ascertain  the  Area  of  an  Ellipse,  Fig.  4. 

Rule. — Multiply  the  diameters  together,  the  product  by 
.7854,  and  the  result  will  give  the  area  required. 

Or,  multiply  one  diameter  by  .7854,  and  the  product  by 
the  other. 

Or,  dx^X -7854= arm. 

Example. — The  transverse  diameter  of  an  ellipse,  A  B, 
Fig.  4,  is  12  inches,  and  its  conjugate,  C  D,  9 ;  what  is  its 
area? 

12 X 9  X.7854:= 84. 8232  =product  of  diameters  and  .7 854:= result  re- 
quired. 

Ex.  2.  The  diameters  of  an  ellipse  are  70  and  50  feet ; 
what  is  its  area*?  Ans.  2748.9  feet. 

Centre  of  Gravity.     Is  in  its  geometrical  centre. 


/Segment  of  an  Ellipse. 

To  ascertain  the  Area  of  a  Segment  of  an  Ellipse  when  its  base  is 
parallel  to  either  axis,  as  e  if  Fig.  5.       / 

Rule. — Divide  the  height  of1  the  segment  b  i  by  the  diam- 
eter or  axis,  a  b,  of  which  it  is  a  part,  and  find  in  the  table  of 
areas  of  segments  of  a  circle,  p.  134-138,  a  segment  having 
the  same  versed  sine  as  this  quotient ;  then  multiply  the  area 
of  the  segment  thus  found  and  the  two  axes  of  the  ellipse  to- 
gether, and  the  product  will  give  the  area  required. 

Gr,  h-^-dx  tab.  area  x  d .  d/z=area. 

Fig.  5. 


240  CONIC   SECTIONS. 

Example. — The  height,  b  i,  Fig.  5,  is  5  inches,  and  the 
axes  of  the  ellipse  are  30  and  20 ;  what  is  the  area  of  the 
segment  ? 

^-=.1666  =  tabular  versed  sine,  the  area  of  which  (p.  135)  is  .08554. 
Hence,  .08554  X  30  x20=51.324=*Ae  area  required. 

Ex.  2.  The  height  of  a  segment  at  right  angles  or  perpen- 
dicular to  the  transverse  diameter  of  an  ellipse  is  6.25  inches, 
and  the  diameters  are  16  and  25  %  what  is  the  area  of  the 
segment?  Ans.  61.42  inches. 

Ex.  3.  The  height  of  a  segment  of  an  ellipse  at  right  angles 
to  the  conjugate  diameter  is  25  inches,  and  the  diameters  are 
50  and  70  ;  what  is  the  area  of  the  segment  ? 

Ans.  1374.415  inches. 


The  area  of  an  elliptic  segment  may  also  be  found  by  the  following  rule: 
Ascertain  the  segment  of  the  circle  described  upon  the  same  axis  to 

which  the  base  of  the  segment  is  perpendicular. 

Then,  as  this  axis  is  to  the  other  axis,  so  is  the  circular  segment  to 

the  elliptical  segment. 

Illustration. — In  the  above  example,  the  axis  to  which  the  base  of 
the  segment  is  perpendicular  is  the  conjugate,  50,  and  the  height  of  the 
segment  25.     Also,  the  area  of  the  segment  is  one  half  of  that  of  a  circle 

of  50  ^a^er=1963;4954^981.7472. 

Hence,  50:  70: :  981.75  :  1374.45  =area  of  elliptic  segment. 


PARABOLA. 

1.  To  describe  a  Parabola,  the  Base  and  Height  being  given,  Fig.  6. 

Operation. — Draw  an  isosceles  triangle,  as  A  B  D,  the 
base  of  which  shall  be  equal  to,  and  its  height,  B  c,  twice  that 
of  the  proposed  parabola. 

Divide  each  side,  A  B,  D  B,  into  any  number  of  equal 
parts;  then  draw  lines,  1  1,  2  2,  3  3,  etc.,  and  their  intersec- 
tion will  define  the  curve  of  a  parabola. 


Fig.  6. 


CONIC    SECTIONS. 
15, 


241 


2.  To  describe  a  Parabola,  any  Ordinate  to  the  Axis  and  its  Ab- 
scissa being  given,  Fig.  7. 

Operation. — Bisect  the  ordinate,  as  A  o  in  r;  join  B  r, 
and  draw  r  s  perpendicular  to  it,  meeting  the  axis  continued  to  s. 

Draw  B  c  and  B  n  each  equal  to  o  s,  and  n  will  be  the  fo- 
cus of  the  curve. 

Take  any  number  of  points,  1  1,  etc.,  on  the  axis,  through 
which  draw  the  double  ordinates  2  12,  etc.,  of  an  indefinite 
length.  Then,  with  the  radii  cn,cl,  etc.,  and  focal  centre  n, 
describe  arcs  cutting  the  corresponding  ordinates  in  the  points 
2  2,  etc.,  and  the  curve  ABC,  drawn  through  all  the  points 
of  intersection,  will  define  the  parabola  required. 

Note. — The  line  2  n  2,  passing  through  the  focus  n,  is  the  parameter. 

Fig.  7. 

B. 


242 


CONIC    SECTIONS. 


To  ascertain  either  Ordinate  or  Abscissa  of  a  Parabola,  the  other 
Ordinate  and  the  Abscissa?,  or  the  other  Abscissa  and  the  Or- 
dinates  being  given,  Fig.  8. 
Rule. — As  either  abscissa  is  to  the  square  of  its  ordinate, 

so  is  the  other  abscissa  to  the  square  of  its  ordinate. 


Or,  1.  =  o 


—if*. 


3.  — j^-=a. 


2. 


4. 


o  zxa 


a 


—a' 


Or,  as  the  square  root  of  any  abscissa  is  to  its  ordinate,  so  is 
the  square  root  of  any  other  abscissa  to  its  ordinate. 

„  ox  Va'       / 

Hence,  -. —  =zo  . 

ya 

Fig.  8. 


| 


Example. — The  abscissa  a  b,  of  the  parabola,  Fig.  8,  is  9, 
its  ordinate,  b  c,  6  ;  what  is  the  ordinate  d  e,  the  abscissa  of 
which,  a  d,  is  16? 

Hence,  9 :  62 : :  1 6  :  64,  and  V  64 = 8 = length  of  ordinate  required 
Or,  V  9  :  6 : :  V 16 :  8  =ordinate  as  before. 

Ex.  2.  The  less  abscissa  of  a  parabola  is  16,  its  ordinate 
10 ;  what  is  the  ordinate,  the  abscissa  of  which  is  36  ? 

Ans.  15. 
Ex.  3.  The  abscissae  of  a  parabola  are  9  and  16,  and  their 
I  corresponding  ordinates  6  and  8 ;  any  three  of  these  being 
taken,  it  is  required  to  find  the  fourth. 


CONIC    SECTIONS.    '  243 


1.  62x16^— ^64,  and  t/64: =8= ordinate. 

j  y 

2.  ZJll  __Ar_36,  and  v'36  =  6=Wiwafe. 

3.  • — — — =— —  =  9z=less  abscissa. 

82  64 

.    82x9      576     f.       _    . 

4.  — — -  =— -=16=abscissa. 

6"1  ob 


Parabolic  Curve. 

To  ascertain,  the  Length  of  the  Curve  of  a  Parabola  cut  off  by  a 
Double  Ordinate. 
Rule. — To  the  square  of  the  ordinate  add  ^  of  the  square 
of  the  abscissa,  and  the  square  root  of  this  sum,  multiplied  by 
two,  will  give  the  length  of  the  curve  nearly. . 


4a2\ 
Or,  -y/{o2-\ — —  J  x2  =  length  of  curve. 


Example. — The  ordinate,  d  e,  Fig.  8,  is  8,  and  its  abscis- 
sa, a  d}  16  ;  what  is  the  length  of  the  curve/  a  e  f 

4x  16a 
82  +  — ^ — =105. 333= sum  of  square  of  the  ordinate  and  %  of  the 
o 

square  of  the  abscissa,  and  V  405.333= 20.133,  which  X  2 =10.267= length 
required. 

Note. — This  rule  can  be  used  only  when  the  abscissa  does  not  ex- 
ceed half  the  ordinate.  The  length  of  the  curve  in  other  cases  is  to  be 
found  by  means  of  hyperbolic  logarithms,  as  shown  by  writers  on 
fluxions. 


Ex.  2.  The  ordinate  is  16  inches,  and  its  abscissa  32; 
what  is  the  length  of  the  curve?  Ans.  80.533  inches. 

Ex.  3.  The  abscissa  is  20,  and  its  ordinate  12  inches; 
what  is  the  length  of  the  curve  ?  Ans.  52.05  inches. 

Ex.  4.  The  abscissa  is  5  feet,  and  its  ordinate  3 ;  what  is 
the  length  of  the  curve?  Ans.  13.014  feet. 


244  *    CONIC   SECTIONS. 

Parabola. 

To  ascertain  the  Area  of  a  Parabola,  Fig.  9. 
Rule. — Multiply  the  base  by  the  height  and  J  of  the  prod- 
uct will  give  the  area  required.*  ^ 
Or,  §  bxh=zarea. 

Fig.  9.  b 


Example. — What  is  the  area  of  the  parabola  a  b  c,  Fig.  9, 
the  height,  b  e,  being  16  inches,  and  the  base,  or  double  ordi- 
nate, a  c,  16  ? 

16  X 16  =  256  =product  of  base  and  height,  and  §  of  256  =  170.667  = 
area  required. 

Ex.  2.  The  height  of  an  abscissa  of  a  parabola  is  32  inches, 
and  the  base  or  double  ordinate  is  16;  what  is  its  area? 

Ans.  341.333  square  inches. 

Ex.  3.  The  height  of  a  parabola  is  50  inches,  and  its  base 
24  ;  what  is  its  area  ?  Ans.  800  square  inches. 

To  ascertain  the  Area  of  a  Frustrwn  of  a  Parabola,  Fig.  10. 

Rule. — Multiply  the  difference  of  the  cubes  of  the  two 

ends  of  the  frustrum,  a  b,  c  d,  by  twice  its  altitude,  e  o,  and 

divide  the  product  by  three  times  the  difference  of  the  squares 

of  the  ends. 

d3(^>d/3  x%h 
Or,  -= -q — —=area,  d  and  d/  representing  the  lengths  of 

the  base  and  lesser  end. 

*  Corollary. — A  parabola  is  §  of  its  circumscribing  parallelogram. 


CONIC   SECTIONS.  245 

Fig.  10. 


Example. — The  ends  of  a  frustrum  of  a  parabola,  a  b  and 
c  d,  Fig.  10,  are  10  and  6  inches,  and  the  height,  e  o,  is  10 
inches ;  what  is  its  area  ? 


103co63X  10  x  2  =  15680= difference  of  cubes  of  the  ends  X  twice  the 
height. 

15680-f-102co  62  X  3=81.667  =preceding  product-r-3  times  the  differ- 
ence of  the  squares  of  the  ends —  area  required. 

Ex.  2.  The  base  and  lesser  end  of  a  frustrum  of  a  parabola 
are  30  and  24  inches,  and  the  height  22  inches ;  what  is  its 
area*?  Arts.  596.444  inches. 

Ex.  3.  The  base  and  upper  end  of  a  frustrum  of  a  parabo- 
la are  30  and  20  inches,  and  the  height  20  inches ;  what  is 
the  area"?  Ans.  506.667  inches. 

Ex.  4.  The  ends  of  the  frustrum  of  a  parabola  are  5  and 
4  feet,  and  its  height  2.5  feet ;  what  is  its  area  ? 

Ans.  1 1.296  feet. 

Ex.  5.  The  ends  of  a  frustrum  of  a  parabola  are  8.5  and  7 
feet,  and  its  height  6  feet ;  what  is  its  area *? 

Ans.  46.645  feet. 

Note. — Any  parabolic  frustrum  is  equal  to  a  parabola  of  the  same 
altitude,  the  base  of  which  is  equal  to  the  base  of  the  frustrum,  in- 
creased by  a  third  proportional  to  the  sum  of  the  two  ends  and  the 
lesser  end. 

Illustration. — In  Example  1,  the  base  and  end  are  10  and  6. 

Then,  10  +  6  :  6:  '.6 :  2. 25  =  third  proportional  to  the  sum  of  the  two  ends 
and  the  lesser  end. 

Hence,  10  +  2.25  =  12.25=swn  of  length  of  base  of  parabola  and  third 
proportional,  the  area  of  which,  the  height  being  10,  is=81.667. 


246 


CONIC    SECTIONS. 


HYPERBOLA. 

To  describe  an  Hyperbola,  the  Transverse  and  Conjugate  Diam- 
eters being  given,  Fig.  11. 

Fig.  11. 


/ 

er/_ 

C 

//£     s 

A 
i\  a 

Bj/1 

^    s 

1 
\ 

0 

«v                 iX/ 

f     n        n        n      n 

s 

y 


Let  A  B  represent  the  transverse  diameter,  and  C  D  the 
conjugate. 

Draw  C  e  parallel  to  A  B,  and  e  r  parallel  to  CD;  draw 
o  e,  and  with  the  radius  o  e,  with  o  as  a  centre,  describe  the 
circle  F  e  f  r,  cutting  the  transverse  axis  produced  in  F  and 
/;  then  will  F  and  /  be  the  foci  of  the  figure. 

In  o  B  produced  take  any  number  of  points,  n,  n,  etc., 
and  from  F  and  f  as  centres,  with  A  n  and  B  n  as  radii,  de- 
scribe arcs  cutting  each  other  in  s,  s,  etc.  Through  s,  s,  etc., 
draw  the  curve  s  s  s  B  s  s  s,  and  it  will  be  the  hyperbola  required. 

Note. — If  straight  lines,  as  o  e  y  and  o  r  y}  are  drawn  from  the  cen- 
tre o  through  the  extremities  e  r,  they  will  be  the  asymptotes  of  the 
hyperbola,  the  property  of  which  is  to  approach  continually  to  the  curve, 
and  yet  never  to  touch  it. 

When  the  Foci  and  the  Conjugate  Axis  are  given. 

Let  F  and /be  the  foci,  and  C  D  the  conjugate  axis,  as  in  the  pre- 
ceding figure. 

Through  C  draw  g  C  e  parallel  to  F  and/;  then,  with  o  as  a  centre 
and  oFas  a  radius,  describe  an  arc  cutting  g  C  e  at  g  and  e ;  from 
these  points  let  fall  perpendiculars  upon  the  line  connecting  F  and/  and 
the  part  intercepted  between  them,  as  A  B,  will  be  the  transverse  axis. 


CONIC    SECTIONS. 


247 


To  ascertain  the  Ordinate  of  an  Hyperbola,  the  Transverse  and 
Conjugate  Diameters  and  the  Abscissa  being  given,  Fig.  12. 
Rule. — As  the  transverse  diameter  is  to  the  conjugate,  so 
is  the  square  root  of  the  product  of  the  abscissae  to  the  ordi- 
nate required. 


Or, 


c  X  Va  X  a* 
~~t 


■.ordinate. 


Note. — 1.  In  hyperbolas,  the  less  abscissa,  added  to  the  axis  (the 
transverse  diameter),  gives  the  greater. 

2.  The  difference  of  two  lines  drawn  from  the  foci  of  any  hyperbola 
to  any  point  in  the  curve  is  equal  to  its  transverse  diameter. 

Fig.  12. 


Example. — The  hyperbola,  a  b  c,  Fig.  12,  has  a  transverse 
diameter,  a  t,  of  120  inches,  a  conjugate,  df  of  72,  and  the 
abscissa,  a  e,  is  40 ;  what  is  the  length  of  the  ordinate  e  c? 

40  + 1 20  =  1 60 =sum  of  less  abscissa  and  axis=gveater  abscissa. 
120 :  72: :  -/(40  X  160) :  ±8=ordinate  required. 

Ex.  2.  The  transverse  diameter  of  an  hyperbola  is  25,  the 
conjugate  15,  and  the  less  abscissa  6  inches;  what  is  the 
length  of  the  ordinate?  Ans.  8.1829  inches. 

Ex.  3.  The  transverse  diameter  of  an  hyperbola  is  5  feet, 
the  conjugate  3,  and  the  less  abscissa  1.8  feet ;  what  is  the 
length  of  the  ordinate  ?  Ans.  2  feet. 

Ex.  4.  The  transverse  diameter  of  an  hyperbola  is  5  feet, 
the  conjugate  3,  and  the  greater  abscissa  8.667  feet;  what  is 
the  length  of  the  ordinate  ?  Ans.  3.67  feet. 


248  CONIC    SECTIONS. 

To  ascertain  the  Abscissa,  the  Transverse  and  Conjugate  Diameters 

and  the  Ordinate  being  given,  Fig.  12. 

Eule. — As  the  conjugate  diameter  is  to  the  transverse,  so 

is  the  square  root  of  the  sum  of  the  squares  of  the  ordinate 

and  semi-conjugate  to  the  distance  between  the  ordinate  and 

the  centre,  or  half  the  sum  of  the  abscissas. 

-  Then,  the  sum  of  this  distance  and  the  semi-transverse  will 

give  the  greater  abscissa,  and  their  difference  the  less  abscissa. 

*-v/o2  +  (c~2)2     a+a'     .   7/.  7  „  ,      ,     . 

Or, s= — <r — =z  half  the  sum  of  the  abscissa. 

a-\-a'     t  n  a-\-a'     t 

Example. — The  transverse  diameter,  a  t,  of  an  hyperbola, 
Fig.  12,  is  120  inches,  the  conjugate,  d  f,  72,  and  the  ordi- 
nate, e  c,  48  ;■  what  are  the  lengths  of  the  abscissas,  t  e  and  ae? 

72 :  120: :  V48a+(72-f-2)a  =  60 :  100 —half the  sum  of  the  abscissce. 

100 +(120-^2)  =  160 =above  sum  added  to  the  semi-transverse  =  the 
greater  abscissa,  and 

100  —  (120-i-2)=40=aJore  sum  subtracted  from  the  semi-transverse^ 
the  less  abscissa. 

Ex.  2.  The  transverse  and  conjugate  diameters  of  an  hy- 
perbola are  25  and  15  inches,  and  the  ordinate  8.1829  ;  what 
are  the  lengths  of  the  abscissae?  Ans.  6  and  31  inches. 

To  ascertain  the  Conjugate  Diameter,  the  Transverse  Diameter, 
the  Abscissce,  and  Ordinate  being  given,  Fig.  12. 
Rule. — As  the  square  root  of  the  product  of  the  abscissas 
is  to  the  ordinate,  so  is  the  transverse  diameter  to  the  conju- 
gate. 

o  X  t 

Or,  —77 77 = conjugate  diameter. 

V(axa  ) 

Example. — The  transverse  diameter,  a  b,  of  an  hyperbo- 
la, Fig.  12,  is  120,  the  ordinate,  e  c,  48,  and  the  abscissas, 
t  e  and  a  e,  160  and  40 ;  what  is  the  length  of  the  conju- 
gate^/?^ ;. 

V  40X160=80 ;  48 : 1 120 :  12=conjugate  required. 


CONIC   SECTIONS.  249 

Ex.  2.  The  transverse  diameter  of  an  hyperbola  is  25 
inches,  the  ordinate  8.1829,  and  the  abscissas  6  and  31; 
what  is  the  length  of  the  conjugate?  Ans.  15  inches. 

To  ascertain  the  Transverse  Diameter,  the  Conjugate,  the  Ordi- 
nate, and  an  Abscissa  being  given,  Fig.  12. 

Eule. — Add  the  square  of  the  ordinate  to  the  square  of  the 
semi-conjugate,  and  extract  the  square  root  of  their  sum. 

Take  the  sum  or  difference  of  the  semi-conjugate  and  this 
root,  according  as  the  greater  or  less  abscissa  is  used.^ 

Then,  as  the  square  of  the  ordinate  is  to  the  product  of  the 
abscissa  and  conjugate,  so  is  the  sum  or  difference  above 
found  to  the  transverse  diameter  required. 

Or,  (a  or  a  x c  x ( Vo2-\-(c-i-2)2 d-c-±-2))-~o2  =  transverse 
diameter. 

Example. — The  conjugate  diameter,  df,  of  an  hyperbola, 
Fig.  12,  is  72  inches,  the  ordinate,  e  c,  48,  and  the  less  ab- 
scissa, a  e,  40 ;  what  is  the  length  of  the  transverse  diam- 
eter, a  t  f 


V  ±8*  + {7 2-r-2)2=Q0= square  root  of  the  squares  of  the  ordinate  and 
semi-conjugate. 

60  +  72-*-2=96=sm??i  of  above  root  and  the  semi-conjugate  (the  less 
abscissa  being  used). 

40  X  72  =2880  =product  of  abscissa  and  conjugate. 

482 :  2880 : :  96 :  120 = transverse  diameter. 

Ex.  2.  The  conjugate  diameter  of  an  hyperbola  is  15 
inches,  the  ordinate  8.1829,  and  the  greater  abscissa  31 ; 
what  is  the  length  of  the  transverse  diameter  ? 

Ans.  25  inches. 
Ex.  3.  The  conjugate  diameter  of  an  hyperbola  is  6  feet  8 
inches,  the  ordinate  4  feet,  and  the  less  abscissa  3  feet  4 
inches*,  what  is  the  length  of  the  transverse  diameter? 

Ans.  142.336  feet. 

*  When  the  greater  abscissa  is  used,  the  difference  is  taken,  and 
contrariwise. 

L  2 


250  CONIC    SECTIONS. 


To  ascertain  the  Length  of  any  Arc  of  an  Hyperbola,  commencing 
at  the  vertex,  Fig.  13. 

Rule. — To  19  times  the  transverse  diameter  add  21  times 
the  parameter  of  the  axis,  and  multiply  the  sum  by  the  quo- 
tient of  the  less  abscissa  divided  by  the  transverse  diameter. 

To  9  times  the  transverse  diameter  add  21  times  the  pa- 
rameter, and  multiply  the  sum  by  the  quotient  of  the  less  ab- 
scissa divided  by  the  transverse  diameter. 

To  each  of  the  products  thus  found  add  15  times  the  pa- 
rameter, and  divide  the  former  by  the  latter ;  then  this  quo- 
tient, being  multiplied  by  the  ordinate,  will  give  the  length  of 
the  arc  nearly. 


tx  19  +  21  xp X-+ 15  xp 

Or, xo—arc  nearly. 

tx   9+21Xi?Xy-fl5xp 

Fig.  13. 


Example. — In  the  hyperbola  a  b  c,  Fig.  13,  the  transverse 
diameter  is  120,  the  conjugate  80,  the  ordinate  e  c  48,  and  the 
less  abscissa,  a  e,  40 ;  required  the  length  of  the  arc  b  a  c? 

120 :  80 : :  80 :  53.3333  ^parameter. 

40 

120 x  19+53.3333 X 21  X—=  1133.3333  =proe&c«  of  the  sum  of  Id 

times  the  transverse  and  21  times  the  parameter,  by  the  quotient  of  the  less 
abscissa  and  transverse. 

Note. — As  the  transverse  diameter  is  to  the  conjugate,  so  is  the  con- 
jugate to  the  parameter.     (See  rule,  p.  231.) 


CONIC    SECTIONS.  251 


40 


120x9+53.3333x21  X-—  =  733. 3333=product  of  the  sum  of  9  times 

the  transverse  and  21  times  the  parameter,  by  the  quotient  of  the  less  abscis- 
sa and  transverse. 


1133.3333 +53.3333  xl5+-(733.3333+53.3333x  15)  =  1.2622  =  quo- 
tient of  former  product  and  15  times  the  parameter  flatter  product  and  15 
times  the  parameter. 

1.2622  X  48  =  60.5856  =above  quotient  X  the  ordinate  =  the  length  re- 
quired. 

Ex.  2.  The  transverse  diameter  of  an  hyperbola  is  60,  the 
conjugate  36,  the  ordinate  24,  and  the  less  abscissa  20 
inches ;  required  the  length  of  the  curve. 

Ans.  31.324  feet. 

Ex.  3.  The  transverse  diameter  of  an  hyperbola  is  25,  the 
conjugate  15,  and  the  less  abscissa  6 ;  required  the  length  of 
the  curve.  Ans.lO.2Hl. 

Ex.  4.  The  transverse  diameter  of  an  hyperbola  is  50,  the 
conjugate  15,  and  the  greater  abscissa  31 ;  what  is  the  length 
of  the  curve?  Ans.5.9395. 

To  ascertain  the  Area  of  an  Hyperbola,  the  Transverse,  Conju- 
gate, and  less  Abscissa  being  given,  Fig.  13. 

Rule. — To  the  product  of  the  transverse  diameter  and  less 
abscissa  add  4  of  the  square  of  this  abscissa,  and  multiply  the 
square  root  of  the  sum  by  21. 

Add  4  times  the  square  root  of  the  product  of  the  trans- 
verse diameter  and  less  abscissa  to  the  product  last  found,  and 
divide  the  sum  by  75. 

Divide  4  times  the  product  of  the  conjugate  diameter  and 
less  abscissa  by  the  transverse  diameter,  and  this  last  quo- 
tient, multiplied  by  the  former,  will  give  the  area  required 
nearly. 

n     V*Xa/+-X2x21+-(\/*Xa/x4)     cxa'x* 

Or, n X  ——=area. 

Example. — The  transverse  diameter  of  an  hyperbola,  Fig. 
13,  is  60  inches,  the  conjugate  36,  and  the  less  abscissa  or 
height,  a  e,  20  ;  what  is  the  area  of  the  figure  ? 


252  CONIC   SECTIONS. 

60x20+f  of  202  =  1485.7143=swm  of  the  product  of  the  transverse 
and  abscissa  and  -f-  of  the  square  of  the  abscissa. 

V  1485.7143  X  21  =  809.424=21  times  the  square  root  of  the  above  sum. 

V  60  X  20  X  4  +  809. 424=947.988  =  sum  of  above  result  and  the  square 
root  of  4  times  the  product  of  the  transverse  and  abscissa. 

947.988-^75  =  12.6398  ^quotient  of  above  result-r-75. 

36  x  20  x  4 

— X  12.6398  =  606.7 104:=product  of  4  times  the  product  of  the 

conjugate  and  abscissa-^-the  transverse  and  the  above  quotient = area  re- 
quired. 

Ex.  2.  The  axes  of  an  hyperbola  are  100  and  60,  and  the 
less  abscissa  50  inches ;  what  is  its  area  ? 

Ans.  22.3636  square  feet. 

Ex.  3.  Required  the  area  of  an  hyperbola,  the  greater  ab- 
scissa being  6.25  feet,  the  ordinate  2.165,  and  the  transverse 
diameter  4.167.  Ans.  9.6008  square  feet. 


CONIC    SECTIONS.  253 


CONICAL  UNGULAS. 

To  ascertain  the  Convex  Surface  of  a  Conic  Ungula,  Fig.  14. 
Fig.  14. 

c  / T~y%   6 


1.  When  the  Section  passes  through  the  opposite  Extremities  of  the  Ends 
of  the  Frustrum,  as  a  e. 

Rule. — Let  d  represent  diameter  of  greater  end,  d'  diameter  of  less  end, 
and  h  the  perpendicular  height  of  the  frustrum,  as  o  r. 

Then,  '^^VW+(d^dyx d* (.d+d^xdd0 -convex surface of 

elliptic  ungula,  a  eh.  -v 

Example. — The  diameters  ab,ce,  of  the  frustrum  of  the  cone,  Fig. 
14,  are  10  and  5  inches,  and  the  height,  o  r,  10;  what  is  the  convex 
surface  of  the  elliptic  ungula  a  eh? 

.7854     .7854 


d-d'     10-5 


.15708,   and  .15708  X  V 4  h*  +  (d-d')2  =  .15708  X 


V(d+dr+d<f) 


•/(4x  100+10-5  =3.2383,  and 3.2383  xda-     v        a  J =3.2383 

X100~ 
required. 


X 100 (10+5  X10X5)=32383 x iQM7=zl52mS2=convexsurface 


To  ascertain  the  Contents  of  a  Conic  Ungula,  Fig.  14. 
Rule. — Let  n=.7854. 


,™       d2—d'Vdd'     ndh  „  .  „.    .  7  , 

lnen, X  — — = contents  of  the  greater  elliptic  ungula  a  e  b. 

dVdd'-d'*     nd'h  '*         ...    .  . 

-j — -s; —  X— - — ^contents  of  the  less  elliptic  ungula  ace. 

(d%-d'%y     nh      „_, 

3 —  X  — = difference  of  these  ungnlas. 


254  CONIC    SECTIONS. 

Example. — The  diameters,  height,  and  section  of  the  frustrum  being 
as  in  the  preceding  example,  what  are  its  contents  ? 

102-5\/10x5       100-5X7.071      100-35.355       __       ,., 

ttt-t = - = =  12.920,  which   X 

10  —  5  5  5 

<1?=78^4_26        and  12.929x26.18=338.4812=co»ten<* 


3 

required. 


To  ascertain  the  Convex  Surface  of  a  Conic  Ungula,  Fig.  15. 
Fig.  15. 


2.  When  the  Section  cuts  off"  part  of  the  Base,  and  makes  the  Angle  erb 
less  than  the  angle  cab. 
Note.— When  a  Segment  greater  than  a  Semicircle  is  to  he  found, 
Subtract  the  quotient  of  ita  veraed  sine,  divided  by  the  diameter  of  the  circle,  from 
unity.  Then  take  the  tabular  segment  (p.  134-137)  which  corresponds  to  the  re- 
mainder from  .78539,  the  whole  tabular  area,  and  the  remainder  will  be  the  tabular 
segment  corresponding  to  the  segment  required. 

RULE._  _Ly 4 h*+(d-dy  x (tab.  seg.  ibs~x lx(^,}"ar 
d—d'  ..-.,.  d2  d'—ar 

X  V-p X  area  of  segment  of  circle   a  b,  the  height  of  which  is 

d  X  — — —  J  ==  convex  surface  of  elliptic  ungula  iebs. 

Example. — The  diameters  ab,  ce,  of  the  frustrum  of  the  cone,  Fig. 
15,  are  10  and  5  inches,  and  the  height  o  n  10;  what  is  the  convex 
surface  of  the  elliptic  ungula,  the  base  b  r  being  6  inches  ? 

^-^=—^=.2,  and  .2  X  V±h2  +  (d-dy  =  .2  xV(ix  100+To^53) 

=  4.1231. 

Tab.  seg.  i  b  s=-fo=.6,  and.G  — 1  =  .4,  and  tab.  seg.  of  .4  X  102=29.337, 
which -78.539 =49.202. 

d'*_  25    ^X(<J+Q-ar_^X(10  +  5)-4_3.5 

da~100*         d'-ar        ~         5-4  "  1 

vtLL_=v««2Am. 

d—ar         5  —  4     1 


CONIC    SECTIONS. 


255 


Area  of  seg.  cir.  a  b  .  height=dx — — —  =  10x  — —  =2,  and  2-MO 

=  .2,  tab.  seg.  =.11823,  which  X  102=1 1.1823. 

Then,  4. 1231 X  (49.202-^x3.5  X  2.4495  X  11.1823)  =  104.0464  = 
convex  surface  required. 

To  ascertain  the  Contents  of  a  Conic  Ungula,  Fig.  15. 
Rule. — Let  S=tabular  segment  qfbr,  the  versed  sine  being-^-a  b  ;  s= 
tabular  segment,  the  versed  sine  being  b  r  —  (d—d')-r-d'. 
b  r  br 


Then^SXcP-sXd^X 


br       \       Ik 

,         J  X  ,      7=  contents  of 
—d—d  J     d—d 


br  —  d—d'     br- 
the  ungula  ieb  s. 

Example. — The  dimensions  of  the  frustrum  and  ungula  being  the 
same  as  in  the  preceding  example,  what  are  its  contents  ? 

S=-r=— ,  and  ttl^.G  .'.  1  — .6  =  .4,  the  tabular  segment  of  which  is 

(page  135)  .29337,  and  .78539 -.29337 =.49202. 

6—5 
s=br—(d—d'-7-d')=  =  .2,  the  tabular  segment  of  which  is  .11182. 

S  X  d3  =  .  49202  X102 =492.02. 

br  br  


Xd'3X 


IV 


-=.11182  X  53  X  6  X  2.4495=205.427, 


br—d—d'     br—d—d' 
which-492.02=286.593. 

_|&      3J533  _  ^  and  2g6  593  x  >6667  =  19L0715  =  resuh  n_ 

d—d         5 
quired. 

To  ascertain  the  Convex  Surface  of  a  Conic  Ungula,  Fig.  16. 

Fin.  10. 

C  k J — +  d 


3.  When  the  Section  is  parallel  to  one  of  the  Sides  of  the  Frustrum,  as  d  r. 

-rl-pV4:h>+(d-dy  X  area  of  seg.  i  b  «- f|  {d-d)  X  Vd'x(d-d')) 
= convex  surface  of  parabolic  ungula  idb  s. 


256  CONIC    SECTIONS. 

Example. — The  diameters  ab,  cd,  of  the  frustrum  of  the  cone,  Fig. 
16,  are  10  and  5  inches,  and  the  height  or  10;  what  is  the  convex  sur- 
face of  the  ungula  ? 

-1— =— — =  .2,  and  .2x  V  4  h*+(d-dy  =  V  ±x  100  +  (10-5)2  = 
d—d      10  —  5 
.2X20.6155=4.1231. 

Area  ofseg.  i  b  s.-Q.(d-d')X  Vd'X(d-d'j)  =39.27- (? (10-5) X 

V5x(10-5))=39.27-16.667  =  22.603. 

Then,  4.1231  x22.603=93.1944=conuex  surface  required. 

To  ascertain  the  Contents  of  a  Conic  Ungula,  Fig.  16. 
Rule. — Let  A=area  of  base  i  b  s. 

Then,  ( - — -  —  - d'V (d— d')X d')\  X-h ^contents  required. 

Example. — The  diameters  of  a  frustrum  of  a  cone,  Fig.  16,  a  b  and 
c  d,  are  10  and  5  inches,  and  the  height  o  r  10;  what  are  the  contents 
of  the  ungula  i  d  b  s? 

A= 102  X  .7854-^2  =39.27. 
Axd    39.27x10    392.70 


d-d'        10-5 


78.54. 


-d'V{d-d')xd'=^5V{iO-b)x 5  =  6.667X5=33.335,    and  33.335 
o  o 

■78.54=45.205,  which  X-  10=45.205  X 3.334  =  150.7135 ^contents  re- 


quired. 


To  ascertain  the  Convex  Surface  of  a  Conic  Ungula,  Fig.  17. 


4.  When  the  Section  cuts  off  part  of  the  Base,  and  makes  the  Angle  drb 
greater  than  the  Angle  cab. 


-±-j,  X  V  ±h*  +  (d-dy  X  (dr.  seg.  i  b  s  -  ^  X  o  r  -  ^{d  -  d') 
d~d  V  *  br-d=d' 

V )  =  convex  surface  of  hyperbolic  ungula  idbs. 

br—d'—d/ 


CONIC    SECTIONS.  257 

Example. — The  dimensions  of  the  frustrum  being  the  same  as  in  the 
preceding  example,  and  the  height  o  n  and  base  b  r  of  the  ungula  be- 
ing 10  and  2.5  inches,  what  are  its  contents  ? 


d-d' 


—  xV4:h2-\-(d-dy  =  .2x  20.6155=4.1231. 


2.5 
And  (cir.  seg.  i  b  s=— =  .25,  tab.  seg.  =  .153546  X  103  =  15.3546. 

cT2_25    br-$(d-d') br  _2.5— frx  (10-5)  _  6.25 

d~100'     br-(d-d'y       br-(d-d')~     2.5-(10-5)     ~~275~ 

V si =i. 

V  2.5 -(5 -10) 

ok      6  2^ 
Then,  4.1231  X  (15.346--—  X-^  X  1)  =  15.346  -  .625  =  14.721, 
100      Z.o 

which  X  4.1231  =  60.6962  =convex  surface  required. 

To  ascertain  the  Contents  of  a  Conic  Ungula,  Fig.  17. 
Let  the  area  of  the  hyperbolic  section  i  d  s=A,  and  the  area  of  the  cir- 
cular seg.  ib  s=a. 

Then,  _/*    ,x(aXd— Ax — -, = contents  of  the  hyperbolic  ungula 

d  —  h  dr 

i  d  b  s. 

Note.— The  transverse  diameter  of  the  hyp.  seg.=  _      ,  the  conjugate  = 

br 

dV-; r, — r— i  an<i  tne  abscissa  =  dr,  from  which  its  area  may  be  found  by  rule, 

a  —  a  — or 

page  251. 

Example. — The  diameters  a  b,  c  d,  of  the  frustrum  of  the  cone,  Fig. 
17,  are  10  and  5  inches,  the  heights  o  n  and  dr,  of  an  hyperbolic  un- 
gula are  10,  and  the  base  b  r  2.5  inches ;  what  are  its  contents  ? 

A  (by  rule,  page  251)=53.675,  a  =  15.355. 

Then,  -i^  =  ?'S 3* = 3.334,  and  («x<2-Ax-^— -^=(15.355  X  10 
.     a  —  n     10  — 10  dr    J 

=  153.55-53.675=99.875,  which X    ~  '   =24.9688,  which  X3.SS4  = 

83.246  =  contents  required. 


258  MENSURATION   OF   REGULAR   POLYGONS. 


REGULAR  POLYGONS. 

(Appendix  to  Rules,  see  Table,  page  261.) 
TO   ASCERTAIN   THE   ELEMENTS   OP   ANY   REGULAR  POLYGON. 

To  ascertain  the  JRadius  of  the  Inscribed  or  Circumscribing  Circles 
of  a  Polygon. 

1.  When  the  Length  of  a  Side  is  given. 

Rule. — Multiply  the  length  of  the  side  by  the  units  in  col- 
umns A  and  B  of  the  following  table  under  the  head  of  the 
element  required. 

Example. — The  length  of  the  side  of  a  square  (tetragon) 
is  2  inches ;  what  are  the  radii  of  its  inscribed  and  circum- 
scribing circles? 

2x  .5  =  1. =product  of  length  of  side  and  tabular  multipliers^  radius  of 
its  inscribed  circle. 

2  X. 70711  =  1.41422  =product  of  length  of  side  and  tabular  mulliplier 
= radius  of  its  circumscribing  circle. 

Ex.  2.  The  length  of  the  side  of  a  hexagon  is  2  inches ; 
what  is  the  radius  of  its  inscribed  circle  ? 

Ans.  1.73204  inches. 

2.  When  the  Area  is  given. 

Rule. — Extract  the  square  root  of  the  area,  and  multiply 
it  by  the  units  in  columns  C  and  D  of  the  following  table  un- 
der the  head  of  the  element  required. 

Example. — The  area  of  a  square  is  4  inches;  what  are 
the  radii  of  its  inscribed  and  circumscribing  circles  ? 

V  4  X  .5  =  1 .  =radius  of  its  inscribed  circle. 

■/4X  .70711  =  1.41422  =  radius  of  its  circumscribing  circle. 

Ex.  2.  The  area  of  a  hexagon  is  64.952  inches;  what  is 
the  radius  of  its  circumscribing  circle?  Ans.  5  inches. 


MENSURATION  OP  REGULAR  POLYGONS.        259 

3.  When  the  Radius  of  the  Circumscribing  Circle  is  given. 
Rule. — Multiply  the  radius  given  by  the  unit  in  the  col- 
umn E  opposite  to  the  figure  for  which  the  radius  is  required. 

Example. — The  radius  of  the  circumscribing  circle  of  a 
square  is  2  inches ;  what  is  the  radius  of  its  inscribed  circle  % 
2  X. 70711  =  1.414:22 =radius  of  its  inscribed  circle, 

Ex.  2.  The  radius  of  the  circumscribing  circle  of  a  hexa- 
gon is  5  inches ;  what  is  the  radius  of  its  inscribed  circle  % 

Ans.  4.3301  inches. 

4.  When  the  Radius  of  the  Inscribed  Circle  is  given. 
Rule. — Multiply  the  radius  given  by  the  unit  in  the  col- 
umn F  opposite  to  the  figure  for  which  the  radius  is  required. 

Example. — The  radius  of  the  inscribed  circle  of  a  square 
is  1.41421  inches;  what  is  the  radius  of  its  circumscribing 
circle  1 

1.41421  X  1.41421 =2 =radius  of  its  circumscribing  circle. 

Ex.  2.  The  radius  of  the  inscribed  circle  of  a  hexagon  is 
.866025  inches;  what  is  the  radius  of  its  circumscribing  cir- 
cle? Ans.  1  inch. 

To  ascertain  the  Area. 

1.  When  the  Radii  of  the  Inscribed  or  Circumscribing  Circles 
are  given. 

Rule. — Square  the  radius  given,  and  multiply  it  by  the 
unit  in  the  columns  G  and  H  of  the  following  table  under  the 
head  of  the  element  given. 

Example. — The  radius  of  the  inscribed  circle  of  a  square 
is  2  inches ;  what  is  its  area  ? 
22X  4= 16 =area. 

Ex.  2.  The  radius  of  the  circumscribing  circle  of  a  hexa- 
gon is  5  inches;  what  is  its  area?  Ans.  64.952  inches. 


260        MENSURATION  OP  REGULAR  POLYGONS. 

3.    When  the  Length  of  the  Side  is  given. 
.  Eule. — Square  the  length  of  the  side,  and  multiply  it  by 
the  unit  in  column  I. 

ExAMPLE.-^-The  length  of  the  side  of  a  square  is  4  inches ; 
what  is  its  area? 
42Xl  =  16 =area. 

Ex.  2.  The  length  of  the  side  of  a  hexagon  is  5  inches ; 
what  is  its  area*?  Ans.  64.952  inches. 

To  ascertain  the  Length  of  a  Side. 

1.  When  the  Radii  of  the  Inscribed  or  Circumscribing  Circles 
are  given. 

Rule. — Multiply  the  radius  given  by  the  unit  in  the  col- 
umns K  and  L  of  the  following  table  under  the  head  of  the 
radius  given. 

Example. — The  radius  of  the  inscribed  circle  of  a  square 
is  2  inches ;  what  is  the  length  of  a  side  % 
2x2=4  =  length  of  side. 

Ex.  2.  The  radius  of  the  circumscribing  circle  of  a  hexa- 
gon is  5  inches  ;  what  is  the  length  of  a  side  ? 

Ans.  5  inches. 

2.  When  the  Area  is  given. 

Rule. — Extract  the  square  root  of  the  area,  and  multiply 
it  by  the  unit  in  column  M  of  the  following  table. 

Example. — The  area  of  a  square  is  16  inches ;  what  is  the 
length  of  a  side  ? 

V 16  X  1=4= length  of  side. 

Ex.  2.  The  area  of  a  hexagon  is  64.952  inches ;  what  is 
the  length  of  a  side?  Ans.  5  inches. 

Note. — For  other  rules  to  ascertain  the  surface  or  linear  edge  of  a 
polygon,  see  page  64. 


MENSURATION  OF  REGULAR  POLYGONS. 


261 


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262  MENSURATION  OF  POLYHEDRONS. 


MENSURATION   OF   REGULAR  BODIES,   OR 
POLYHEDRONS. 

To  ascertain  the  Elements  of  any  Regular  Body,  Figs.  72,  83,  84, 
85,  and  86,  p.  72,  161-164. 

To  ascertain  the  Radius  of  a  Sphere  that  will  Circumscribe  a  given 
Regular  Body,  and  the  Radius  of  one  also  that  may  be  Inscribed 
within  it. 

1.  When  the  Linear  Edge  is  given. 

Rule. — Multiply  the  linear  edge  by  the  multiplier  opposite 
to  the  body  in  the  columns  A  and  B  in  the  following  table, 
under  the  head  of  the  element  required. 

Example. — The  linear  edge  of  a  hexahedron  or  cube  is  2 
inches ;  what  are  the  radii  of  the  circumscribing  and  inscribed 
spheres  ? 

2  X. 86602  =  1. 73204  =product  of  edge  and  tabular  multiplier = radius 
of  circumscribing  sphere. 

2  X  .5  =  l=product  of  edge  and  tabular  multiplier — radius  of  inscribed 
sphere. 

Ex.  2.  The  linear  edge  of  a  hexahedron  is  10  inches; 
what  are  the  radii  of  its  inscribed  and  circumscribing  spheres  ? 

Ans.  5  and  8.6602  inches, 

2.  When  the  Surface  is  given. 

Rule. — Multiply  the  square  root  of  the  surface  by  the  mul- 
tiplier opposite  to  the  body  in  the  columns  C  and  D  in  the 
following  table,  under  the  head  of  the  element  required. 

Example. — The  surface  of  a  hexahedron  is  25  inches ; 
what  are  the  radii  of  the  circumscribing  and  inscribed  spheres'? 

V25  X  .35355  =  1.76775  =  radius  of  circumscribing  sphere. 
V2S  x  .20412  =  1. 0206  =  radius  of  inscribed  sphere. 


MENSURATION  OF  POLYHEDRONS.  263 

Ex.  2.  The  surface  of  a  hexahedron  is  24  inches ;  what  is 
the  radius  of  its  inscribed  sphere  ?  Arts.  1  inch. 

3.  When  the  Volume  is  given. 

Rule. — Multiply  the  cube  root  of  the  volume  by  the  mul- 
tiplier opposite  to  the  body  in  the  columns  E  and  F  in  the 
following  table  under  the  head  of  the  element  required. 

Example. — The  volume  of  a  hexahedron  is  8  inches ;  what 
are  the  radii  of  its  circumscribing  and  inscribed  spheres  ! 

V  8  X  .86602  =  1.73204:= radius  of  circumscribing  sphere, 
y/  8  X  .5 =1  =  radius  of  inscribed  sphere. 

4.  When  one  of  the  Radii  of  the  Circumscribing  or  Inscribed 
Sphere  alone  is  required^  the  other  being  given. 

Rule. — Multiply  the  given  radius  by  the  multiplier  oppo- 
site to  the  body  in  the  columns  G  and  H  in  the  following  ta- 
ble, under  the  head  of  the  other  radius. 

Example. — The  radius  of  the  circumscribing  sphere  of  a 
hexahedron  is  1  inch;  what  is  the  radius  of  its  inscribed 
sphere  ? 

1  X  .57735 =.57735  = radius  required. 

Ex.  2.  The  radius  of  the  inscribed  sphere  of  a  hexahedron 
is  2  inches ;  what  is  the  radius  of  its  circumscribing  sphere  % 

Ans.  3.4641  inches. 

To  ascertain  the  Linear  Edge  of  a  Polyhedron. 
li  When  the  Radius  of  the  Circumscribing  or  Inscribed  Sphere 
is  given. 

Rule. — Multiply  the  radius  given  by  the  multiplier  oppo- 
site to  the  body  in  the  columns  I  and  K  in  the  following  table. 

Example. — The  radius  of  the  circumscribing  sphere  of  a 
hexahedron  is  1  inch ;  what  is  its  linear  edge  1 

1  X  1.1547  =  1. 1547  =edge  required. 


264  MENSURATION  OF  POLYHEDRONS. 

Ex.  2.  The  radius  of  the  inscribed  sphere  of  a  hexahedron 
is  .57735  inch ;  what  is  its  linear  edge  ? 

Ans.  1.1547  inch. 

2.  When  the  Surface  is  given. 

Rule. — Multiply  the  square  root  of  the  surface  by  the  mul- 
tiplier opposite  to  the  body  in  the  column  L  in  the  following 
table. 

Example. — The  surface  of  a  hexahedron  is  6  inches ;  what 
is  its  linear  edge  ? 

V  6  X  .40825  =  1  =linear  edge. 

Ex.  2.  The  surface  of  a  hexahedron  is  24  inches ;  what  is 
its  linear  edge?  Ans.  2  inches. 

3.  When  the  Volume  is  given. 

Rule. — Multiply  the  cube  root  of  the  volume  by  the  mul- 
tiplier opposite  to  the  body  in  the  column  M  in  the  following 
table. 

Example. — The  volume  of  a  hexahedron  is  .3535  inch; 
what  is  its  linear  edge  ? 

v7 .3535  X  1  =  .7071=Unear  edge. 

Ex.  2.  The  volume  of  a  hexahedron  is  8  inches ;  what  is 
its  linear  edge"?  Ans.  2  inches. 

To  ascertain  the  Surface  of  a  Polyhedron. 
1.    When  the  Radius  of  the  Circumscribing  Sphere  is  given. 

Rule. — Multiply  the  square  of  the  radius  by  the  multi- 
plier opposite  to  the  body  in  the  column  N  in  the  following 
table. 

Example. — The  radius  of  the  circumscribing  sphere  of  a 
hexahedron  is  .866025  inch  ;  what  is  its  surface? 

.86602a  X  8=6= surface  required. 


MENSURATION  OF  POLYHEDRONS.  265 

2.  When  the  Radius  of  the  Inscribed  Sphere  is  given. 

Rule. — Multiply  the  square  of  the  radius  by  the  multi- 
plier opposite  to  the  body  in  the  column  O  in  the  following 
table. 

Example. — The  radius  of  the  inscribed  sphere  of  a  hexa- 
hedron is  .5  inch ;  what  is  its  surface  ? 
.52X  24  =  6= surface  required. 

3.  When  the  Linear  Edge  is  given. 

Rule. — Multiply  the  square  of  the  edge  by  the  multiplier 
opposite  to  the  body  in  the  column  P  in  the  following  table. 

Example. — The  linear  edge  of  a  hexahedron  is  1  inch ; 
what  is  its  surface  ? 

I2  X  6  =  6  =  surface  required. 

4.  When  the  Volume  is  given. 

Rule. — Extract  the  cube  root  of  the  volume,  and  multiply 
the  square  of  the  root  by  the  multiplier  opposite  to  the  body 
in  column  Q  in  the  following  table. 

Example. — The  volume  of  a  hexahedron  is  8  inches ;  what 
is  its  surface  ? 

^8  =  2,  and  22  X  6 = 24  = surface  required. 

Ex.  2.  The  volume  of  a  hexahedron  is  353.5533  inches; 
what  is  its  surface?  Ahs.  300  inches. 

To  ascertain  the  Volume  of  a  Polyhedron. 
1.    When  the  Linear  Edge  is  given. 

Rule. — Cube  the  linear  edge,  and  multiply  it  by  the  multi- 
plier opposite  to  the  body  in  column  R  in  the  following  table. 

Example. — The  linear  edge  of  a  hexahedron  is  2  inches ; 
what  is  its  volume  ? 

23  Xl  =  8=volume  required. 

M 


266  MENSURATION  OF  POLYHEDRONS. 

2.  When  the  Radius  of  the  Circumscribing  Sphere  is  given. 

Rule. — Multiply  the  cube  of  the  radius  given  by  the  mul- 
tiplier opposite  to  the  body  in  the  column  S  in  the  following 
table. 

Example. — The  radius  of  the  circumscribing  sphere  of  a 
hexahedron  is  .5  inch ;  what  is  the  volume  of  the  figure"? 
53  X  1.5396 =.1925  cubic  inch. 

Ex.  2.  The  radius  of  the  circumscribing  sphere  of  a  hexa- 
hedron is  1.73205  inch;  what  is  the  volume  of  the  figure? 

Ans.  8  cubic  inches. 

3.  When  the  Radius  of  the  Inscribed  Sphere  is  given. 

Rule. — Multiply  the  cube  of  the  radius  given  by  the  mul- 
tiplier opposite  to  the  body  in  the  column  T  in  the  following 
table. 

Example. — The  radius  of  the  inscribed  sphere  of  a  hexa- 
hedron is  .5  inch  ;  what  is  its  volume  ? 

.53x8  =  1  cubic  inch. 

Ex.  2.  The  radius  of  the  inscribed  sphere  of  a  hexahedron 
is  3.535  inches ;  what  is  its  volume? 

Ans.  353.3932  cubic  inches. 

4.  When  the  Surface  is  given. 

Rule. — Cube  the  surface  given,  extract  the  square  root, 
and  multiply  the  root  by  the  multiplier  opposite  to  the  body 
in  the  column  U  in  the  following  table. 

Example. — The  surface  of  a  hexahedron  is  6  inches ;  what 
is  its  volume  ? 

VQ3X. 06804=1  cubic  inch. 

Ex.  2.  The  surface  of  a  hexahedron  is  24  inches ;  what  is 
its  volume?  Ans.  8  inches. 

Ex.  3.  The  surface  of  an  octahedron  is  125  inches;  what 
is  its  volume?  Ans.  102.1743  cubic  inches. 


MENSURATION    OP   POLYHEDRONS. 


267 


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268     orophoids  {Domes,  Arched  and  Vaulted  Roofs,  etc.). 


OROPHOIDS*  {Domes,  Arched  and  Vaulted  Roofs,  etc.). 

Definition.  Figures  generated  by  a  curved  line  running  from 
the  periphery  or  the  angles  alone  of  their  base  to  a  common  cen- 
tre at  the  top.  When  the  curve  runs  from  the  periphery  alone, 
and  their  point  of  connection  is  in  the  centre,  they  are  Regular ; 
when  the  point  of  connection  is  eccentric,  they  are  Oblique  ;  ichen 
curves  run  both  from  the  angles  and  some  intermediate  point,  or 
there  is  any  combination  of  elements  in  their  generation,  they  are 
Compound  or  Irregular. 

Orophoids  or  Arched  Roofs  are  either  Domes,  Saloons, 
Vaults,  or  Groins. 

A  Dome  is  formed  by  arches  springing  from  a  circular  or 
polygonal  base,  and  meeting  in  the  centre  at  the  vertex. 

A  Saloon  (frustrum  of  an  orophoid)  is  formed  by  arches 
springing  from  a  circular  or  polygonal  base,  and  connecting 
with  a  flat  roof  or  ceiling  in  the  middle. 

A  Vault  is  formed  by  arches  springing  from  two  opposite 
bases  alone,  and  meeting  in  a  line  at  the  vertex. 

A  Groin]  is  formed  by  the  intersection  of  one  vault  with  an- 
other at  any  angle. 

Orophoids  are  of  the  following  forms : 

1.  Circular  Dome,  Saloon,  or  Vault,  when  generated  by  an 
arc  of  a  circle. 

2.  Elliptical  Dome,  Saloon,  or  Vault,  when  generated  by  an 
arc  of  an  ellipse. 

*  From  opodog,  a  roof.  The  absence  of  any  generic  term  to  denote 
this  class  of  figures  has  induced  the  adoption  of  the  above  term. 

t  The  curved  surface  between  two  adjacent  groins  is  termed  the  sec- 
troid. 


orophoids  (Domes,  Arched  and  Vaulted  Roofs,  etc.).     269 

3.  Gothic  Dome,  Saloon,  or  Vault,  when  generated  by  two 
circular  or  elliptical  arcs  meeting  in  the  centre  of  the  arch. 

4.  Curvilinear  Dome,  Saloon,  or  Vault,  when  generated  by 
an  irregular  curve. 

To  ascertain  the  External  or  Internal  Surface  of  a  Spherical 
Dome.* 

Rule. — Multiply  the  area  of  the  base  by  2,  and  the  prod- 
uct will  give  the  surface  required. 

Or,  a  x  2  —surface. 

Example. — The  external  diameter  of  a  hemispherical  dome 
is  20  feet ;  what  is  the  surface  of  it  ? 

202X  .7854  =314. 16 =area  of  base. 

314.16  x2=628.S2  =  twice  the  area  of  the  base=surface  required. 

Ex.  2.  The  sides  of  a  quadrangular  spherical  dome  are  20 
feet;  what  is  the  surface  of  it?  Ans.  800  square  feet. 

Ex.  3.  The  internal  sides  of  a  hexagonal  spherical  dome 
are  10  feet ;  what  is  the  surface  of  if? 

Ans.  519.62  square  feet. 

To  ascertain  the  Surface  of  a  Side  of  a  Polygonal  Spherical 

Dome. 
Rule. — Multiply  the  base  by  its  length,  divide  by  1.5708, 
and  the  quotient  will  give  the  surface  required. 

#)n  —surface- 

'  1.5708         y  / 

Example. — The  side  of  a  quadrangular  spherical  dome  is 
20  feet ;  what  is  the  surface  of  it  ? 

*  An  elliptical  dome  or  vault  is  either  a  semi-spheroid  (ellipsoid)  or 
a  segment  of  a  spheroid ;  for  rules  to  ascertain  the  surface  of  these,  see 
p.  91-93,  and  125. 

A  parabolic  dome  or  vault  is  a  paraboloid,  or  a  segment  of  one ;  for 
rules  to  ascertain  the  surface  of  which,  see  page  126. 

A  Gothic  dome  or  vault  is  a  semi-circular  spindle,  or  a  segment  of 
one;  for  rules  to  ascertain  the  surface  of  which,  see  p.  12CF-122. 


270     orophoids  (Domes,  Arched  and  Vaulted  Hoofs,  etc.). 

2(K-2  X  3.1416=31.416-^2  =  15.708  =  radius  of  dome  X  3.1416-^2  = 
perimeter  or  length  of  side. 

20  X  15.708=314.16=/>rocto  of  length  of  base  and  length  of  side. 

314.16  -M.5708  =200  =  above  product  -f- 1.5708=  surface  of  side  re- 
quired. 

Ex.  2.  The  side  of  a  hexagonal  spherical  dome  is  10  feet ; 
what  is  the  surface  of  it  ?  Ans.  86.6025  feet. 

To  ascertain  the  Volume  of  a  Spherical  Dome* 
Rule. — Multiply  the  area  of  its  base  by  §  of  the  height, 
and  the  product  will  give  the  volume  required. 
Or,  ax%h=volume. 

Example. — The  diameter  of  the  base  of  a  spherical  dome 
is  20  feet ;  what  is  its  volume? 

202x. 7854=314. 16=«rai  of  base. 

314.16x|  of  -^  =  2094.4 =areaxf  of  height =volwne  required. 

Ex.  2.  The  side  of  a  hexagonal  spherical  dome  is  10  feet ; 
what  is  its  volume?  Ans.  1732.0506  cubic  feet. 

To  ascertain  the  External  or  Internal  Surface  of  a  Saloon. 
Rule. — To  the  area  of  the  ceiling  add  the  surface  of  its 
sides,  and  the  sum  will  give  the  surface  required. 
Or,  a+s=surface. 

Example. — A  saloon  roof  has  a  quadrangular  arch  of  2 
feet  radius  springing  over  a  rectangular  base  of  20  by  16 
feet ;  what  is  its  surface  ? 

20— 2 -{-2  =  16= length  of  ceiling. 
16—2+2  =  12  =width  of  ceiling. 
16  X  12  =  192  =area  of  ceiling. 

— __ • 

*  An  elliptic  dome  is  either  a  semi-spheroid  (ellipsoid)  or  a  segment 
of  a  spheroid.  For  rules  to  ascertain  the  volume  of  them,  see  p. 
179-181. 

Parabolic  and  hyperbolic  domes  are  paraboloids  and  hyperboloids  of 
revolution.    For  rules  to  ascertain  the  volumes  of  them,  see  p.  203-206. 

A  Gothic  dome  is  a  semi-circular  spindle  or  a  segment  of  one.  For 
rules  to  ascertain  the  volume  of  them,  see  p.  188-191. 


OROPHOIds  (Domes,  Arched  and  Vaulted  Roofs,  etc.).     271 

2+2  X  3. 1416  =  12. 5664= circumference  of  arch  having  a  radius  of  2 
feet. 

12.5664-f-4=3.1416  =  /e«s*A  of  arch  at  sides. 

3.1416  X  16  + 16  + 12 +  12  =  175.9296  =  /en#*A  of  arch  X  length  of  sides 
and  ends. 

2+2  X  4 

z=4:^=sum  of  sides  of  mitred  ends  of  sides  of  saloon-^-4:=side  of 

4 
a  quadrangular  dome. 

4x4x2=32=area  of  base  X  2  =  surf ace  of  quadrangular  spherical 
dome.  , 

192  +  175.9296  +  32=399.9296=s«r/ace  of  saloon  required. 

Ex.  2.  A  saloon  roof  has  a  hexagonal  arch  of  4.33  feet  ra- 
dius springing  over  an  oblong  hexagonal  base  having  sides  of 
20  feet  and  ends  of  10  feet ;  what  is  its  surface? 

Ans.  556.5832  square  feet. 

To  ascertain  the  Volume  of  a  Saloon. 

Rule. — Multiply  the  square  of  twice  the  height  of  the  arc 
by  .7854,  divide  the  product  by  4,  and  multiply  the  quotient 
by  the  length  of  the  sides  of  the  ceiling. 

Subtract  the  side  of  the  ceiling  from  the  side  of  the  saloon, 
ascertain  the  area  of  a  like  figure  having  a  side  equal  to  this 
remainder  (see  rule,  p.  60),  and  multiply  this  area  by  two  thirds 
of  the  height  of  ceiling. 

Multiply  the  area  of  the  ceiling  by  the  height  of  it;  and 
this  product,  added  to  the  preceding,  will  give  the  volume  re- 
quired. 

2/i2x.7854 


Or, x  l-\-  %a'-\-a  x  h= volume. 

Example. — The  sides  of  a  quadrangular  circular  saloon 
roof  are  20  feet,  and  the  height  of  the  ceiling  2  feet ;  what  is 
the  volume  of  it  ?  • 

Note. —  When  the  Saloon  is  a  Circle. 

To  ascertain  the  Volume  of  the  Ring  projecting  beyond  the  Diameter  of 
the  Ceiling. 

Ascertain  the  centre  of  gravity  of  a  section  of  the  ring,  and  multi- 
ply the  area  of  this  section  by  the  circumference  described  by  its  centre 
of  gravity.     (See  Example  2,  page  272.) 


272     okophoids  (Domes,  Arched  and  Vaulted  Roofs,  etc.). 

2 

2  X  2  X  .785i-7-4:=3.l4:lG=07ie  fourth  of  product  of  the  square  of  twice 
the  height  of  the  arc  and  .7854. 

3.1416  X  (20-2  +  2)  X  4 =201.0624  =pn>ofac*  of  preceding  quotient  and 
the  length  of  the  sides  of  the  ceiling. 

2+2—20=16,  and  16  —  20=1=  diameter  of  ceiling  subtracted  from 
diameter  of  saloon. 

42X  1  =  16  =product  of  square  of  side  and  tabular  multiplier = area  of 
figure.  ■"_ 

16  X§  of  2 =21.333  =product  of  area  of  base  and  §  of  height  of  ceiling. 

162X  2  =512  =product  of  area  of  ceiling  and  its  height. 

512+21.333+201.0624  =  734.3954=sm7;j  of  above  products =volume 
required. 

Ex.  2.  A  circular  room  40  feet  diameter,  and  25  feet  high 
to  the  ceiling,  is  covered  with  a  saloon  having  a  circular  arch 
of  5  feet  radius ;  required  the  contents  of  the  room  in  cubic 
feet.  Ans.  307 '7 9 .453  feet. 

Operation.— Area  of  40 X 25-5  =  1256.64 X 20=25132.8  =  volume  of 
body  of  room. 

Area  of  flat  portion  of  ceiling,  40-5  +  5  =  706.86x25  co  20  =  3534.3 
= volume  of  body  of  roof  . 

Area  of  quadrant  of  a  circle  having  a  radius  of  5= — - —  =  19.635. 

Volume  of  quadrantal  ring  of  5  feet  height  and  base  =area  thereof 
X  the  circumference  described  by  its  centre  of  gravity.     (See  p.  209.) 

Hence  (by  rule,  p.  84),  the  centre  of  gravity  of  the  ring  having  the  sec- 
tion of  a  sector  of  a  circle  of  5  feet  radius  is  3.001  feet  from  the  angle  of  it. 

Then  (by  rule,  p.  55),  the  hypothenuse  of  the  right  angle  (3)  being  alone 
given,  the  length  of  the  side  (that  is,  the  distance  of  the  centre  of  gravity 
of  the  sector  from  the  vertical  side  ofit)=2.122.  THxerefore,  40s—  5  +  5 
+2.122  X  2 =34.244  =diameter  of  circle  described  by  centre  of  gravity  of 
quadrantal  ring. 

Consequently,  circumference  of  34.244  X  19.635  =2112. 353 =volume  of 
quadrantal  ring. 

Volume  of  body  of  room,  25132.8 
"  "  roof,      3534.3 

"  ring,      2112.353 

30779.453  cubic  feet = volume  required. 

Ex.  3.  A  quadrangular  building  having  sides  of  40  and  30 
feet  is  covered  with  a  saloon  25  feet  in  height  from  the  floor, 
having  an  arch  of  5  feet  radius ;  what  is  the  volume  of  the 
saloon?  Ans.  29296.83  cubic  feet. 


orophoids  (Domes,  Arched  and  Vaulted  Hoofs,  etc.).     273 

To  ascertain  the  Surface  of  a  Vault. 
Eule. — Multiply  the  length  of  the  arch  by  the  length  of 
the  vault,  and  the  product  will  give  the  surface  required. 
Or,  px  l=zsurface,p  representing  the  perimeter  of  the  arch. 

Example. — What  is  the  concave  or  internal  surface  of  a 
circular  vault,  the  width  of  it  being  40  feet  and  the  length  8(^| 

40  X  3.1416-^2  X  80 =5026.56  =product  of  length  of  arch  and  length  of 
vault=result  required. 

Ex.  2.  The  width  of  an  elliptic  arched  vault  is  18  feet,  its 
height  12,  and  its  length  50 ;  what  is  its  internal  surface*? 

Ans.  1666.085  square  feet. 

To  ascertain  the  Volume  of  a  Vault. 
Eule.— Multiply  the  area  of  a  section  of  the  vault  by  its 
length,  and  the  product  will  give  the  volume  required. 
Or,  a  x  I— volume. 

Example. — The  width  of  a  semi-circular  arched  vault  is 
10  feet,  and  its  length  60 ;  what  is  its  volume? 

102  X  .7854-^2 =39.27  =area  of  semicircle  of  10  feet  span. 

39.27  X  60 =235 6. 2  =product  of  area  and  length=volume  required. 

Ex.  2.  The  width  of  a  semi-elliptic  arched  vault  is  25  feet, 
its  height  17.5,  and  its  length  40  feet ;  what  are  its  contents'? 

Ans.  13744.5  cubic  feet. 

To  ascertain  the  Internal  ffurface  of  a  Circular  Groin. 
Eule. — Multiply  the  area  of  the  base  by  1.1416,  and  the 
product  will  give  the  surface  required. 
Or,  a  x  1.1416  ^surface. 

Note. — The  exact  surface  or  volume  of  a  groin  is  obtained  by  subtract- 
ing from  the  sum  of  tfce  surfaces  or  volumes  of  the  two  vaults  composing 
the  groin,  the  surface  or  volume  of  the  quadrantal  arch  formed  by  them. 

Thus,  in  Example  1,  the  surface  of  a  circular  groin  of  12  feet  base  is 
as  follows : 

Surface  of  vault,  12  X  3.1416 -f- 2  =  18.8496,  which  X  12  and  2  = 
452. 3904=  sur/ ace  of  the  two  vaults. 

122  X  2 =288 =surface  of  the  quadrantal  arch. 

Hence,  452.3904-288=  164.3904  =surface. 
M2 


274     orophoids  (Domes,  Arched  and  Vaulted  Roofs,  etc.). 

Example. — What  is  the  surface  of  a  circular  groin,  a  side 
of  its  square  being  12  feet. 

122X  1. 1416 ^1G4. 390 i=  product  of  area  of  base  and  1.14 16 ^surface 
required. 

Ex.  2.  What  is  the  surface  of  a  circular  groin,  a  side  of 

its  square  being  8  feet?  Ans.  73.0624  feet. 

•■%' 

Note. — This  rule  may  be  observed  in  elliptical  groins,  as  the  error 
or  difference  is  too  small  to  be  regarded  in  ordinary  practice. 

To  ascertain  the  Volume  of  a  Circular  or  an  Elliptical  Groin. 

Rule. — Multiply  the  area  of  the  base  by  the  height,  and 
the  product  again  by  .904,  and  it  will  give  the  volume  re- 
quired. 

Or,  axhx .904c=:volwne. 

Example. — What  is  the  volume  of  the  vacuity  or  space 
formed  by  a  circular  groin,  one  side  of  its  square  being  10 
feet? 

1 02X  5  X. 904=  452  =product  of  area  of  base,  the  height  and  .904  = 
volume  required. 

Ex.  2.  What  is  the  volume  of  the  vacuity  formed  by  an 
elliptic  groin,  one  side  of  its  square  being  24  feet,  and  its 
height  9  feet?  Ans.  4686.336  cubic  feet. 

To  ascertain  the  Internal  Surface  of  a  Triangular  Groin. 

Rule. — Ascertain  the  length  of  a  side  of  the  arch,  multi- 
ply it  by  twice  the  width  of  the  vault,  and  the  product  will 
give  the  surface  required.* 

Or,  lxbx2=^surface. 

Example. — The  width  of  a  triangular  groin  is  12  feet,  and 
its  height  12  ;  what  is  its  internal  surface? 


-/(122  +  12H-2  )  =  13AlG4:=length  of  one  side  of  arch. 
13.4164x12x2=321. 9936=product  of  length  of  side  and  twice  the 
base=result  required. 

*  See  note,  page  273. 


orophoids  (Domes,  Arched  and  Vaulted  Roofs,  etc.).     275 


To  ascertain  the  Volume  of  the  Materials  that  form  the  Groin. 
Rule. — Multiply  the  area  of  the  base  by  the  height,  inclu- 
ding the  work  to  the  top  of  the  groin,  and  from  this  product 
subtract  the  volume  of  the  vacuity ;  the  result  will  give  the 
volume  required. 

A  General  Rule  for  the  Measurement  of  the  Contents  of  Arches 
is  thus : 

From  the  volume  of  the  whole,  considered  as  a  solid,  from 
the  springing  of  the  arch  to  the  outside  of  it,  deduct  the  vacu- 
ity contained  between  the  said  springing  and  the  under  side 
of  it,  and  the  remainder  will  give  the  contents  of  the  solid 
part. 

In  measuring  works  where  there  are  many  groins  in  a 
range,  the  cylindrical  pieces  between  the  groins,  and  on  their 
sides,  must  be  computed  separately. 

When  the  upper  sides  of  orophoids,  whether  vaults  or  groins,  are 
built  up  solid,  above  the  haunches,  to  the  height  of  the  crown,  it  is 
evident  that  the  product  of  the  area  of  the  base  and  the  height  will 
be  the  whole  'contents.  And  for  the  volume  of  the  vacuity  to  be  de- 
ducted, take  the  area  of  its  base,  computing  its  mean  height  according 
to  its  figure. 


276  .  BOARD    AND   TIMBER   MEASURE. 


BOARD  AND  TIMBER  MEASURE. 

To  ascertain  the  Surface  of  a  Board  or  Plank. 

Rule. — Multiply  the  length  by  the  breadth,  and  the  prod- 
uct will  give  the  surface  required. 

Or,  lxb=surface. 

Note. — When  the  piece  is  tapering,  add  the  breadths  of  the  two  ends 
together,  and  take  half  the  sum  for  the  mean  breadth. 

Example. — The  length  of  a  plank  is  16  feet,  and  its  breadth 
15  inches;  what  is  its  surface  ?. 

16  X  1.25(15)  =20=product  of  length  and  breadth = surface  required. 

Ex.  2.  The  length  of  a  plank  is  25  feet,  and  its  breadth 
14  inches  ;  what  is  its  surface?       Am.,  29.167  square  feet. 

Ex.  3.  The  length  of  a  plank  is  18  feet,  and  its  widths  at 
the  ends  are  17  and  19  inches;  what  is  its  surface? 

Ans.  27  square  feet. 

To  ascertain  the  Contents  of  Squared  Timber. 
Rule. — Multiply  the  breadth  by  the  thickness,  and  this 
product  by  the  length,  and  it  will  give  the  contents  required. 
Or,  bxtxl=contents. 

Example. — The  length  of  a  piece  of  square  timber  is  20 
feet,  its  sides  at  its  less  end  are  15  inches,  and  at  its  greater 
end  19  ;  what  are  its  contents'? 

19  +  15^2  =  17,  and  172X20-M44  =  40.1388 ^product  of  square  of 
mean  side  and  the  kngth-^-lH  to  produce  feet — contents  required. 

Ex.  2.  The  ends  of  a  piece  of  timber  are  18  and  22  inches 
square,  and  the  length  of  it  is  22.5  feet ;  what  are  its  con- 
tents? Ans.  62.5  cubic  feet. 

Note. — 1.  If  the  piece  tapers  regularly  from  one  end  to  the  other,  the 
breadth  and  thickness,  taken  in  the  middle,  will  be  the  mean  breadth 
and  thickness. 

2.  If  the  piece  does  not  taper  regularly,  but  is  thicker  in  some  places 
than  in  others,  take  several  different  dimensions,  and  their  sum,  divided 
by  the  number  of  them,  will  give  the  mean  dimensions. 


BOARD   AND   TIMBER   MEASURE.  277 

To  ascertain  the  Contents  of  Round  or  Unsquared  Timber. 

Eule. — Multiply  the  square  of  one  fifth  of  the  girth  by 
twice  the  length,  and  the  product  will  give  the  contents 
nearly.*      2 

Or,  c-=-5  x  2l=z  contents. 

Example. — The  diameter  of  a  round  piece  of  timber  is 
23^  inches,  and  its  length  18  feet ;  what  are  its  contents'? 


75  (circumference  of  23-|)-f-5  =  15,  and  152=225,  and  225x18x2 
=8100,  which-j-144 =56.25  cubic  feet. f 

Ex.  2.  Xhe  circumference  of  a  round  piece  of  timber  is  168 
inches,  and  its  length  15  ;  what  are  its  contents  % 

Ans.  235.2  cubic  feet. 

*  The  ordinary  rule  is  to  square  one  fourth  of  the  girth,  and  multi- 
ply it  by -the  length. 

In  order  to  show  the  fallacy  of  taking  one  fourth  of  the  girth  for  the 
side  of  a  mean  square,  take  the  following  example  : 

A  diameter  of  13.5  inches  will  give  an  area  of  143.13.  Hence,  a 
piece  of  round  timber  of  this  diameter  will  have  nearly  a  square  foot  of 
area  of  section.  j. 

The  circumference  for  a  diameter  of  13.5=42.41,  and  42.41-7-4  = 
10.6,  and  10.62  =  112.36  =area  of  section  of  timber. 

By  the  above  rule,  42.41-7-5=8.48,  and  8.482  =  71.91,  which  X  2  = 
143.82  =area  of  section  of  timber. 

t  When  feet  are  multiplied  by  inches,  -=-144  to  obtain  cubic  feet. 


278 


APPENDIX  TO  MENSURATION  OF  SURFACES. 


MENSURATION  OF  SURFACES. 

To  ascertain  the  Length  of  an' Elliptic  Curve  which  is  less  than 
half  of  the  entire  Figure,  Fig.  1. 
Fig.  1.  > 


Geometrically. — Let  the  curve  of  which  the  length  is  re- 
quired be  a  b  c. 

Extend  the  versed  sine  b  d  to  meet  the  centre  of  the  curve  in  e. 

Draw  the  line  c  e,  and  from  e,  with  the  distance  e  b,  de- 
scribe b  h  ;  bisect  c  h  in  a,  and  from  e,  with  the  radius  e  i,  de- 
scribe k  i,  and  it  is  equal  to  half  the  arc  ab  c. 

To  ascertain  the  Length  when  the  Curve  is  greater  than  half  the 
entire  Figure. 
Rule. — Find  by  the  above  problem  the  curve  of  the  less 
portion  of  the  figure,  and  subtract  it  from  the  circumference 
of  the  ellipse,  and  the  remainder  will  be  the  length  of  the 
curve  required. 

Parabolic  Spindle. 
To  ascertain  the  Surface  of  a  Parabolic  Spindle*  Fig.  2. 
Fig.  2. 


*  By  Professor  G.  B.  Docharty,  New  York  Free  Academy. 


APPENDIX  TO  MENSURATION  OF  SURFACES.      279 

Let  o  T>=Pf  A  o—q, 

a  o  =    p  r    =x,  and 
ap  =z    or    —y. 
Then,  from  the  nature  of  the  curve, 

p  r2  :  A  o2  ::  Dr:Dc; 
that  is,  x2:q2:\  p—y.p. 

p  x2  =  q2  p—q2  y / 
q2  y~p(q2— X2); 

And,  by  differentiating, 

_  2pxdx         _    _     4»2a;2cZa:2 

<*2/= *--= — ;  .\dy2=-+—- . 

q2  Q 

Let  d  s  represent  the  differential  of  the  surface ;  then, 

ds=2  ny\/dx2-\-dy2. 

Substitute  for  y  its  value,  ^  (q2—x2),  and  for  dy2  its  value, 


2 

ldx2 

,  we  have 

9 


4:p2x2dx2 

~* ,  we  have 

rt4  ' 


^JL(tf.-Xi)\J dx 


^      A  27TPS    1  iVi      4    A4  +  4j02a.2 

Or,ds= — f-  (q2—  x2)  dxy  ±— — | 

=  <^^(q2-x2)dx(q*+4p2x2f 
=^dx(q^+4p2x2f-?-^x2(q^4p2x2fdx; 
.\s=^jqodx(q^4p2x2f-^^CqoX2dx(q^4p2x2f 

=rP  {Jfe4+^2^)i+41°g-^+^^T4^^)-^ 

Wp2q2-q*  q^\ 

32  p*  g'2pS 


280  APPENDIX    TO    MENSURATION    OP    SURFACES. 

Which  is  the  formula  for  the  surface  of  one  half  of  the  spin- 
dle, orADc. 

The  logarithms  indicated  in  the  preceding  and  subsequent  formula 
are  hyperbolical  or  Naperian. 

If  a  table  of  them  is  not  at  hand,  one  of  common  logarithms  may 
be  used  instead,  by  dividing  the  common  logarithm  by  .4343,  and 
using  the  quotient  when  the  logarithm  is  indicated  to  be  used  in  the 
formula. 

To  ascertain  the  Surface  of  a  Frustrum  of  a  Circular  Spindle, 
Fig.  3'. 


L i I 

a  d 

Let  R  represent  the  radius  of  the  circle,  an  arc  of  which 
generates  the  spindle,  as  ad;  D  the  distance  of  the  centre 
of  the  spindle  from  the  centre  of  the  circle,  asao;  d  and  d' 
the  distances  of  the  two  bases  from  the  centre  of  the  spindle, 
as  o  c,  os;  h—d/^pd,  the  altitude,  c  s,  of  the  frustrum,  or  the 
distance  between  the  bases. 

d  d/ 

Find  z  and  zf  from  the  formulae  sin.  z——,  sin.  zf ——. 

AC  xC 

Find  Z  from  the  formula  Z— —  . 

Find  the  surface  from  the  formula  S=2  7rR(A— D  Z). 

In  the  formulae  for  h  and  z,  the  upper  sign  is  used  when  both 
bases  are  on  the  same  side  of  the  centre  of  the  spindle ;  the 
lower  sign  when  the  centre  is  between  them.  Z  is  the  arc 
which  generates  the  frustrum  expressed  in  units  of  the  radius. 

Corollary. — The  entire  surface  of  the  spindle  may  be  found 
from  the  formula  S  =  2  7r  R  (I— D  z),  I  being  the  length  of  the 

spindle,  and  z  being  determined  from  sin.  £*=—  multiplied 
by  ttt^,  and  D  from  D^Rcos.  \z. 


APPENDIX   TO   MENSURATION    OF    SURFACES. 


281 


CENTRES  OF  GRAVITY  OF  SURFACES. 

Semi- Spheroid  or  Ellipsoid,  Fig,  4. 

Fig.  4.  g 


Prolate. 


f»3 y  3 

Distance  from  C—       ^    ,  in  which  C  d,  £fc  radius  of  the 


auxiliary  circle,  =zr=- 


a- 


-.,  a  and  b  representing  the  semi- 


Va2-b2 

transverse  and  conjugate  axes,  y=.ef,  and  S=ara«  of  segment  of 
plane  C  g  ef. 

Oblate.  = — _= 5 2-V , 

|(&vV2  +  &2-r/2log.r'+r/2log.(\/r/2  +  62  +  &)) 

£2 

in  which  r'=  _,  5  awd  a  representing  the  semi-conjugate 

Va2  —  b2 
and  transverse  axes. 


Segment  of  Semi- Spheroid  or  Ellipsoid,  Figs,  5  and  6. 

Fig.  5. 
Prolate. 

A 


282      APPENDIX  TO  MENSURATION  OP  SURFACES. 

3 


Distance  from  C: 


(r2_/2)7_(ra_   2V-8T  y3-y'3    . 


ivhich  y=g  s,y' '—e  o,  and  S—area  of  plane  s  g  e  o,  /=Cs. 

Oblate. 

/ 


Distance  from  C  — 


r'*  +  b^-(r'*  +  P)% 


(V 


§(bVr'  *  +  bi+r'2log.Vr'*  +  b*  +  b-l^r'2  +  l2-r'*log.(Vr,*  +  l2+l)) 

b2 
in  which  r'= —  ,  l=Cs. 


Va2-b* 


Surface  of  a  Frustrum  of  a  Circular  Sjpindle, 
Fig.  7,  p.  283. 

r2 r'2 

The  distance  from  the  centre  of  the  spindle =—— — =r — -,  r 

z{h — D.z) 

and  r'  being  the  radii  of  the  two  bases,  e  and  s;  h  the  distance 
between  the  two  bases ;  D  the  distance  of  the  centre  of  the  spindle 
from  the  centre  of  the  circle,  as  a  o ;  z  the  generating  arc,  ex- 
pressed in  units  of  the  radius. 


'Fig.  7, 


APPENDIX   TO    MENSURATION    OF    SURFACES.  283 

Surface  of  a  Segment  of  a  Circular  Spindle,  as  b  c, 
Fig.  8. 

Fig.  8. 

t 
/ 
/ 


g  h         a  d 

r2 
Distance  from  centre  of  spindle =  r— ; — =r — r,  r  representing 

2(h— D.z) 

the  radius  of  the  base ;  the  other  symbols  the  same  as  given  on 

page  282. 

Note. — This  last  formula  is  essentially  the  same  as  the  follow- 
ing. 


Segment  of  a  Circular  Spindle. 
Distance  from  centre =- 


-\q  -1A~] 

2  \_g  —  I  —  a  (sin. sin.     -  J 


_1 9  -1 1 

sin.      -  and  sin.      -  denoting  the  arcs,  the  sines  of  which  are 
r  r 

7 

respectively  -  and  -  ;  g=fo,  l=zoc,  r— radius  of  circle— ad,  a 
z=:ao=z  Vr2—g2,  and  b  — radius  of  end  circle  of  segment 


Paraboloid  of  Revolution. 

(See  Fig.  65,  on  page  126.) 

tv  *         c                       1     CJp2  +  J2)*(8ft2-2pa)  +  2jp6 
Distance  from  vertex———  xr  T — '— - - — *  '^    r  , 

b2 
a — altitude,  b  —  radius  of  base,  and  p = — . 


284 


APPENDIX   TO   MENSURATION    OF    SOLIDS. 


MENSURATION  OF  SOLIDS. 

Cycloidal  Swindle. 

To  ascertain  the  Contents  of  the  Frustrum  of  a  Cycloidal  Spin- 
dle, b  eco,  Fig.  9. 

Fig.  9. 


e 

~>N 

L 

f 

iiiiiiiii!! 

1> 

|((2r- Z)^(2p+5^r+15Z^r2-15r3ver.sin.  ^+15^r3)) 

dc 

—contents,  I  representing  ef,  r——,  and  p,  as  before,  3.1416, 

m 

~H  .        / 

ver.  sin.     — ,  symbol  for  the  arc,  the  v.  s.  of which =—. 


To  ascertain  the  Contents  of  the  Segment  of  a  Cycloidal  Spindle, 
a  be,  Fig.  9. 


(l5  r3 ver. sin.    '--(2 r-$ (2  P  +  5 fir  + 15 /*r2)) 


P 

6 
tents. 


con- 


\  % 


APPENDIX   TO   MENSURATION    OF    SURFACES.  285 

Frustra  of  Spheroids,  or  Ellipsoids  of  Revolution, 
e  c  df  Figs.  104  and  104*  page  182. 

Distance  from  centre  of  spheroid. 

_    J4      3d(2a*-d*)    „tJ  4      3d(2P-o?) 

Prolate^    g  ^^    ;  0»A=i___J;  arepresent- 

ing  the  semi-transverse  axis,  h  the  semi-conjugate,  and  d  the  height 
of  the  frustrum. 


286 


APPENDIX   TO    MENSURATION    OF    SOLIDS. 


SOLIDS  OF  REVOLUTION. 

To  ascertain  the  Volume  of  a  Solid  of  Revolution*  Fig.  7. 
Fig.  7. 


Let  A  X,  the  axis  of  x,  be  the  axis  of  revolution,  and  c  m  d 
the  generating  curve,  A  n=:r,  m  n=y,  the  co-ordinates  of  any 
point  of  the  curve,  and  let  the  solid  be  terminated  by  planes 
perpendicular  to  the  axis,  cutting  it  at  o  and  s. 

Let  Ao=a,  and  As=b,  the  abscissae  of  these  points;  o  c 
— r,  and  s  d=r',  the  radii  of  the  two  bases.  The  origin,  A, 
may  be  taken  at  any  convenient  point  on  A  X. 

The  general  formula,!  when  A  X  is  the  axis  of  revolution,  is 

V=zpfy2  d  xy  in  which  p  —  3. 1416  ;  f  is  the  symbol  of  integra- 
tion, and  d  that  of  the  differential. 

If,  in  the  expression  for  V,  y2  or  d  x  be  eliminated  by  means 
of  the  equation  of  the  generating  curve,  and  the  integration  be 
effected  between  the  limits  x=a  and  x— b,  or  y=r  and  y=r', 
the  value  of  V  is  determined. 

Corollary. — If  A  Y,  or  the  axis  of  y,  is  the  axis  of  revolu- 
tion, then, 

Y=pfx2dy, 
which  differs  from  the  preceding  simply  by  the  interchange 
of  the  letters  x  and  y. 

*  By  Professor  J.  H.  C.  Coffin,  U.  S.  N. 

f  This  formula  is  thus  read :  The  volume  is  equal  to  p  times  the  in- 
tegral of  y-  multiplied  by  the  differential  of  x. 


APPENDIX   TO    MENSURATION    OF    SOLIDS. 


287 


Example.  To  ascertain  the  Volume  of  a  Cylinder,  Fig.  8. 
Fig.  8.  A 


c 

r 

0 

m 

i\ 

r' 

s 

The  generating  line,  c  m  d,  is  a  right  line  parallel  to  the 
axis,  AX;  its  equation  is  y=r=r/. 

Whence  V  z=pr2fdx;  and  integrating  between  x=a  and 
x =  b, 

V=pr2(b—a);  or,  letting  h= b— a(=o  s),  the  altitude  of 
the  cylinder, 

V—pr2h;  or,  since  pr2  = the  area  of  the  base, 

The  volume  of  a  cylinder  is  equal  to  the  area  of  its  base  multi- 
plied by  its  altitude. 

Example  2.  To  ascertain  the  Volume  of  the  Frustrum  of  a 
Cone  with  a  Circular  Base. 

The  generating  curve  is  a  straight  line  inclined  to  the  axis. 
Let  h=:b— a(  —  o  s),  the  altitude  of  the  frustrum.  The  equa- 
tion of  the  generating  line  is  (the  origin  being  at  the  vertex), 

y  = — r —  x,  whence  dx=— dy,  and 

h  r  — r 

V a= /  y2dy;  and  integrating  between  y = r  and y — r ', 

ph(r/3~r2)  . 

Y=z     3(r/-^~;  orreducing> 

V=^hp(r2-±-r'2-\-rr');  i.  e.  (since  pr2  and  pr2  are  the 
two  bases,  and  p  r  r'  is  a  mean  proportional  between  the  two), 

The  volume  of  such  a  conical  frustrum  is  equal  to  the  sum  of  the 
two  bases  and  their  mean  proportional,  multiplied  by  one  third  of 
the  altitude. 


288 


APPENDIX   TO   MENSURATION    OF    SOLIDS. 


Or  the  following  rule  may  be  used : 

Add  together  the  squares  of  the  radii  of  the  two  bases  and  the 
product  of  those  radii,  and  multiply  the  sum  by  one  third  of  the 
altitude  and  the  number  3.1416. 

Consequently, if  the  volume  of  the  cone  itself  be  required, 
r'—o,  and  V— \hpr2 ;  that  is, 

The  volume  of  a  cone  is  equal  to  its  base  multiplied  by  one 
third  of  its  altitude. 


Example  3.  To  ascertain  the  Volume  of  a  Spherical  Segment, 
Fig.  9. 

Fig.  9.  A 


\ 


The  generating  curve,  c  m  d,  is  an  arc  of  a  circle,  and  its 
equation,  if  the  origin  be  taken  at  the  centre  of  the  sphere,  C,  is 

y2=~R2  — x2  (R  being  the  radius  of  the  sphere);  whence, 

V =pf(R2 —x2) d x;  and  integrating  between  x—a  andx—b, 

V=p[R2(b  — a)— i(b3— a3)]  ;  or,  letting  h=b—a,  the  al- 
titude of  the  segment,  and  substituting  for  R2  its  value, 
W2+r'2)+\{a2  +  b2), 

V=hp  fa  (r2  +  r/2)  +  i  h2~\ ;  that  is, 

The  volume  of  a  spherical  segment  is  equal  to  half  the  sum  of 
its  bases -\- one  sixth  the  area  of  the  circle,  the  radius  of  which  is 
equal  to  the  altitude,  multiplied  by  the  altitude. 

Or  the  following  rule  may  be  used : 

Add  together  the  square  of  the  altitude  and  three  times  the 
squares  of  the  radii  of  the  two  bases,  and  multiply  the  sum  by  one 
sixth  of  the  altitude  and  by  the  number  3.1416. 


APPENDIX   TO   MENSURATION   OP   SOLIDS.  289 

Corollary  1.  If  one  of  the  bases=0,  V=£  hp  (3  r2  +  h2)  ; 
that  is, 

The  volume  of  a  spherical  segment  with  a  single  base  is  equal  to 
the  sum  of  the  area  of  a  circle,  the  radius  of  which  is  the  alti- 
tude, and  three  times  the  base,  multiplied  by  one  sixth  the  alti- 
tude. > 

2.  If  both  bases =0,  the  segment  becomes  the  entire  sphere, 
and 

V=^p  h3 ;  or,  since  A=2E,  the  diameter, 

V=fpR2x^=f^R3;  that  is, 

The  volume  of  a  sphere  is  equal  to  two  thirds  of  a  cylinder  leav- 
ing the  same  diameter  and  altitude. 

Or,  It  is  equal  to  four  thirds  of  the  cube  of  its  radius,  multi- 
plied by  3.1416. 

2.  If  both  bases =0,  the  segment  becomes  the  entire  sphere, 
and 

V=^p  h3 ;  or,  since  ^2E=D,  the  diameter, 

V=:%p~R2xh=%pTL3=ipT>3;  that  is, 

T/ie  volume  of  a  sphere  is  equal  to  two  thirds  of  a  cylinder  hav- 
ing the  same  diameter  and  altitude. 

Or,  It  is  equal  to  four  thirds  of  the  cube  of  its  radius,  multi- 
plied by  3.1416. 

Or,  It  is  equal  to  one  sixth  of  the  cube  of  its  diameter,  multiplied 
by  3.1416. 

Example  4.  To  ascertain  the  Volume  of  a  Segment  of  a  Pro- 
late Spheroid. 

The  generating  curve,  c  m  d,  is  an  arc  of  an  ellipse ;  and 
if  the   origin  be  taken   at  the   centre,  C,  the   equation  is 

B2 

y2=.-—  (A2— x2),  in  which 

A 

A  and  B  are  respectively  the  semi-transverse  and  semi-con- 
jugate axes  of  the  ellipse.      Whence, 

B2  r 

Y=zp—  I  (A2— x2)d  x;   and   integrating   between  x= a 

and  x=zb, 

N 


290       APPENDIX  TO  MENSURATION  OF  SOLIDS. 

B2 

Y=p  —  [A2(b-a)-i(P-a3y];or,lettiiagh=b-a(=os), 
.A. 

the  altitude  of  the  segment,  substituting  for  A2  its  value, 
A2 
2  "r2  (r2+r/2)+i  («2+&2)>  an<*  reducing, 

B2  h2~\ 
Y=hp[^(r2+r^)+  — .-J;  that  is, 

The  volume  of  such  a  segment  is  equal  to  half  the  sum  of  the  bases 
4-  one  sixth  of  the  area  of  the  circle  the  radius  of  which  is  the  alti- 

/B'2\ 

tude,  multiplied  by  (-Tg  J>  the  square  of  the  ratio  of  the  axes  of 

the  generating  ellipse,  multiplied  by  the  altitude. 

Or  the  following  rule  may  be  used : 

Multiply  the  squares  of  the  altitude  and  semi-conjugate  axis  to- 
gether ;  divide  by  the  square  of  the  semi-transverse  axis ;  add  to- 
gether the  quotient  and  three  times  the  squares  of  the  radii  of  the 
two  bases ;  and  multiply  the  sum  by  one  sixth  of  the  altitude  and 
by  the  number  3.1416. 

Corollary  1.  If  one  of  the  bases =0,  the  expression  for  the 


[r2      B2  h2~\ 
"2+A2'"6_r 


2.  If  both  bases  =:0,  the  segment  is  the  entire  spheroid,  and 
the  altitude  A=2A;  and 

V=;?B2x^;  or, 

The  volume  of  such  a  spheroid  is  equal  to  -|  the  volume  of  a 
cylinder  of  the  same  altitude,  the  base  of  which  is  equal  to  the  mid- 
dle section  of  the  spheroid. 

Example  5.  To  ascertain  the  Solidity  of  a  Segment  of  an  Ob- 
late Spheroid. 

A2 
The  equation  is  y2=—  (B2— x2) ;  whence 

A2    C 
V—p^   l(B2—x2)dx;  and  integrating  between  x—a 

and  x—b,  and  reducing,  as  in  Example  4, 

v=AKiO-2+>-'2)+p.J} 


APPENDIX   TO   MENSURATION    OF    SOLIDS.  291 

Example  6.  To  ascertain  the  Volume  of  a  Segment  of  an  Hy- 
perboloid  of  Revolution. 

B2 

The  equation  is  y2z=—(x2— A2) ;  whence 
A. 

B2   C  • 

V=p-r^  I  (x2—A2)dx;  and  integrating  and  reducing,  as 

in  Examples  4  and  5, 

B2  h2~\ 
V=Ai»[i(r»+r")-5;  -gj- 

Example  7.  To  ascertain  the  Volume  of  a  Segment  of  a  Par- 
aboloid. 

The  equation  is  y2  =  2  ~Px,  in  which  2  P=the  parameter. 

V=2  Fpfxdx;  or,  integrating  between  x=.a  and  x—b, 

V^~Pp(b2  —  a2);  or,  since  r2=2P«,  and  r/2=2PJ,  and 
h~b— a, 

Y=^hp(r2+r'2). 


Another  method  of  determining  the  volume  of  a  solid  of 
revolution  is  to  ascertain  the  area  of  the  generating  surface 
and  the  distance  of  its  centre  of  gravity  from  the  axis  of  rev- 
olution. 

Let  A=zthe  area  osdc (Fig.  7). 

g—the  distance  of  its  centre  of  gravity,  G,  from  the  axis 

AX;  then  +, 

Y=z2pgxA;  that  is, 
The  volume  is  equal  to'  the  area  of  the  generating  surface  mul- 
tiplied by  the  circumference  of  the  circle  described  by  its  centre  of 
gravity. 


292  APPENDIX  TO   MENSURATION   OF    SOLIDS. 


CENTRES  OF  GRAVITY. 

To  ascertain  the  Centre  of  Gravity  of  a  Solid  of  Revolution. 

The  centre  of  gravity  is  upon  the  axis  of  revolution,  and 
it  is  necessary  to  determine  only  its  distance  from  some  par- 
ticular point,  as,  for  example,  the  vertex,  or  the  intersection 
of  the  axis  by  a  base,  or  the  origin  of  co-ordinates,  A,  Fig.  7. 

As  before,  Let  AX,  the  axis  ofx,  be  the  axis  of  revolution,  etc. 
(See  page  286.) 

Let  G/  be  the  centre  of  gravity  of  the  solid,  and  AGc'=g. 

The  general  formula  is  ■  _ 

pfy2xdx     fy2xdx 

9=z        V        =  fy2dx' 

If  y  or  x  and  d  x  be  eliminated  by  means  of  the  equation 
of  the  curve  which  generates  the  surface,  and  the  integration 
be  effected  between  the  limits  x=a  and  x=b,  or  y=r  and 
y—rf,  the  distance  of  the  centre  of  gravity,  G',  from  the  ori- 
gin of  co-ordinates  is  determined. 

Example.  To  ascertain  the  Centre  of  Gravity  of  a  Cylinder 

with  a  Circular  Base.     (See  Fig.  8,  page  287.) 

The  equation  is  y=r=r/  ;  whence 

r  2  I'x  d  x 
<7= — J- — - — ,  and  integrating  between  the  limits  x=a  and 

x=b, 

b2—a2 
g=^-rr -—^(b-\-a)  ;  or,  if  the  origin  be  taken  at  o, 

g=^b  =  half  the  altitude. 

Example  2.  To  ascertain  the  Centre  of  Gravity  of  the  Frus- 
trum  of  a  Cone  with  a  Circular  Base.    (See  Example  2,  p.  287.) 

The  equation  is  y— — ; —  x;  whence  x=—z y,  and  d  x 

h  r  —r" 

=-7 dy;  and 


APPENDIX  TO   MENSURATION   OP    SOLIDS.  293 

fij3dy 


and  integrating, 


r'—r' fy2dy 

3  h  (V/4 r4) 

g-  — —, /3  ;  and  reducing, 

*     4(r  —  r)(r3— H) 

3ft(r/a  +  r2)(r/+r) 

0-  4(r^3_r3) 

IPtoce,  To  ascertain  the  distance  of  the  centre  of  gravity  from 
the  pertex  of  the  cone, 

Multiply  together  the  sum  of  the  radii  of  the  two  bases,  the 
sum  of  their  squares,  and  f  the  altitude,  and  divide  the  product  by 
the  difference  of  the  cubes  of  those  radii.     . 

Corollary  1.  If  the  distance  from  the  greater  base  (or  5G') 
be  required, 

r/4_4/r3  +  3r4  r/24-2rr'  +  3r2 


_x;   (r/+r)a  +  2f8 
-4/i'(r/+r)2_rr/' 

2.  For  the  entire  cone,  r=0,  and 

<7=-f  ^,  or  g'=\h;  that  is, 
T^e  distance  of  the  centre  of  gravity  from  the  vertex  is  equal  to 
2-  the  altitude  ;  or,  from  the  base,  is  equal  to  J  the  altitude. 

Example  3.  To  ascertain  the  Centre  of  Gravity  of  a  Spher- 
ical Segment.     (See  Example  3,  page  288.) 

Equation  of  curve,  y2=~R2— x2  ;  whence  xdx=  —ydy;  and 
pf—y3dy  r'4 — r4  y/4_r4 

9=~     ~~V         =4^(r/2+r2)+^2]  =  2%/2  +  r2+^2]; 

Hence,  To  ascertain  the  distance  of  the  centre  of  gravity  of  the 
segment  from  the  centre  of  the  sphere, 

Take  the  difference  of  the  Aih  powers  of  the  radii  of  the  bases 
as  a  dividend;  and  for  the  divisor,  multiply  the  sum  of  the  squares 
of  the  radii  and  ^  the  square  of  the  altitude  by  twice  the  altitude ; 
the  quotient  is  the  distance  required. 

The  centre  of  gravity  is  between  the  centre  of  the  sphere 
and  the  lesser  base. 


294  APPENDIX   TO   MENSUBATION    OP    SOLIDS. 

Corollary.  For  a  segment  with  a  single  base,  r=0 ;  and 

,•*  (R—lh)2 

<7—  — fT3— "^5 — TT~>  ^e  distance  from  the  centre  of 

£Ae  sphere.  " 


Example  4.  To  ascertain  the  Centre  of  Gravity  of  a  Frustrum 
of  a  Prolate  or  Oblate  Spheroid. 

The  distance  from  the  centre  of  the  spheroid  is 

A2 

(r'*—rA) 

n(d±d/)(2A2-d/2-d2)  p 

T*2  J/>_j/,7  3T-   i0r    a 


••5 


2A(r  2+r2+^-— A2) 

prolate  spheroid. 
B2 

A2V  ; ^d±:d')(2B2-d"2-d2)  r 

9~9i,<'^_  2^iAV~*  3W-d'*  +  d'd-d*    t0ran 

oblate  spheroid,  A  cmd  B  representing  semi-transverse  axes,  d 
and  a"  respectively  the  distances  from  the  centre  of  the  spheroid  to 
the  base  and  end  of  the  frustrum.  If  both  these  are  on  the  same 
side  of  the  centre,  the  upper  sign  is  used,  but  if  they  are  on  differ- 
ent sides,  the  lower  sign  is  used. 

Example  5.  To  ascertain  the  Cenfre  of  Gravity  of  a  Segment 
of  a  Prolate  or  Oblate  Spheroid. 


(j± ±  h)2 

<7=^-r — t"T~j  f°r  a  Pr°late  spheroid. 

g=-^5 — TT"?  tor  an  oblate  spheroid. 

-fc> Tf  tl 


Frustrum  of  Hyjperboloid  of  Revolution. 

^.  *  „         11M     Ad'+d){d'2+d2-2a2) 

Distance  trom.centre  of  hyperboloid=:f  — ^ — -,-         —     2  , 

a— semi-transverse  axis,  d— distance  from  centre  to  base  of  seg* 
ment. 


APPENDIX   TO   MENSURATION    OF    SOLIDS.  295 

Segment  of  Hyperboloid  of  Revolution,  Fig.  12,  p.  247. 
Distance  from  centre  of  hyperboloid  (point  of  intersection 

of  the  diameters  t  a  and  d  f)=%  — -,  a  and  das  before, 

and  d/ —distance  from  centre  to  base  of  frustrum. 

Frustrum  of  a  Paraboloid  of  Revolution. 

Distance  from   vertex   of  paraboloid =§ —, — ,  d 

representing  height  of  paraboloid,  and  d/  the  distance  between  the 
frustrum  and  vertex. 


296  APPENDIX  TO  MENSURATION. 

For  the  mensuration  of  Spherical  Triangles  and  Pyra- 
mids, Spirals,  Epi-cycles,  Epi-cycloids,  Cardioids,  Heli- 
coids,  Peli-coids,  etc.,  etc.,  see  Loomis's  Analytical  Geometry 
and  Calculus;  Davies  and  Peck's  Dictionary  of  Mathematics; 
Docharty's  Geometry;  Hackletfs  Geometry. 

For  a  Glossary  and  Explanation  of  Geometrical  Figures, 
see  Davies  and  Peck,  and  the  Library  of  Useful  Knowledge, 
vols.  i.  and  ii. 


carpenters'  slide-rule.  297 


CARPENTERS'  SLIDE-RULE,  OR  GUNTER'S  LINE. 

This  instrument  is  commonly  called  a  Sliding  Rule.  It  is 
constructed  of  two  pieces  of  box- wood  or  ivory,  of  a  foot  in 
length  each,  connected  together  by  a  joint,  which  enables  them 
to  be  folded  up  lengthwise  upon  an  edge. 

On  one  side,  the  whole  rule  is  divided  upon  its  outer  edges 
into  inches  and  eighths,  for  the  purpose  of  taking  dimensions, 
and  upon  its  inner  edges  there  are  several  plane  scales,  each 
divided  into  twelve  parts,  which  are  designed  for  the  reduction 
of  dimensions  for  the  purposes  of  drawing. 

On  the  other  side  there  is  a  metallic  slide,  and  four  lines 
marked  A,  B,  C,  and  D,  the  two  middle,  B  and  C,  being 
upon  the  slide. 

Three  of  these  lines,  A,  B,  and  C,  are  doubled  in  their  di- 
mensions ;  that  is,  they  proceed  from  1  to  10  twice ;  and  the 
fourth  line,  D,  is  a  single  one,  running  from  1  to  10,  and  is 
called  the  line  of  roots* 

The  use  of  the  double  lines,  A  and  B,  is  for  operations  in 
arithmetic,  and  ascertaining  the  areas  of  plane  figures. 

Upon  the  other  part  of  this  face  there  is  a  table  of  gauge 
points  for  the  ascertaining  and  measurement  of  the  different 
elements  and  substances  there  given,  under  the  respective  col- 
umns of  Square,  Circular,  and  Globe ;  and  the  gauge  point  to 
be  selected  for  operation  is  determined  by  the  denomination 
in  which  the  dimensions  are  given ;  thus,  if  in  three  dimen- 
sions, and  all  in  feet,  that  under  F.F.F.  is  to  be  taken ;  if  one 
dimension  is  in  feet  and  the  others  in  inches,  that  under  F.I.I. 
is  to  be  taken ;  and  if  all  are  in  inches  alone,  that  under  I.I.I, 
is  to  be  taken.  Again,  if  of  two  denominations,  or  of  one 
only,  the  gauge  point  is  to  be  taken  from  under  F.I.,  1. 1.,  or 
F.  and  I.,  as  the  case  may  be. 

*  Upon  some  rules,  the  fourth  line,  D,  is  divided  from  4  to  40,  in 
which  case  it  is  termed  a  girth-line,  and  is  then  used  in  the  measure- 
ment of  timber. 

N2 


298  carpenters'  slide-rule. 

The  divisions  on  the  first,  second,  and  third  lines  are  mark- 
ed alike,  each  beginning  at  1,  which  may  be  1,  10,  100,  or 
1000,  etc.,  or  .1,  .01,  .001,  etc. ;  but,  whatever  it  is  assumed 
to  be,  the  second  or  middle  line  of  these  divisions  must  be 
taken  at  10  times  as  many  as  the  first,  and  the  third  line 
must  be  taken  at  10  times  as  many  as  the  second. 

Numeration  is  the  first  operation  to  be  acquired  upon  this 
instrument,  for  when  that  is  understood,  all  other  operations 
will  become  quite  easy.  It  is  first  to  be  observed  that  the 
values  of  the  divisions  upon  the  rule  are  all  arbitrary,  and  the 
value  set  upon  them  must  be  such  as  will  meet  the  require- 
ments of  the  question,  which  must  be  determined  when  a 
question  is  proposed. 

The  principal  divisions  indicated  by  figures  at  1,  2,  3,  4, 
and  so  on  to  10,  are  termed  primes,  and  the  next  divisions 
tenths,  and  these  again  are,  or  may  be,  subdivided  into  hund- 
redth and  thousandth  parts. 

If  the  1  (next  to  the  joint)  represents  one  tenth,  then  will 
the  middle  1  be  a  unit,  or  a  whole  number,  and  the  other 
figures  toward  the  right  hand  are  likewise  whole  numbers, 
from  the  middle  1  to  10  at  the  end;  but  if  the  first  1  rep- 
resents a  unit,  then  the  middle  1  will  be  10,  and  the  10  at 
the  right  hand  100 ;  if  the  first  1  represents  10,  the  middle  1 
will  be  100,  and  that  at  the  right  hand  1000;  always  in- 
creasing in  a  tenfold  proportion,  according  to  the  value  set 
upon  the  first  1.  The  figures  between  1  and  10  are  designated 
after  the  same  manner ;  that  is,  if  1  at  the  beginning  is  one 
tenth,  2  will  be  two  tenths,  and  the  next  2  toward  the  right 
hand  2  units ;  but  if  1  at  the  beginning  is  1  unit,  then  2  will 
be  2  units,  and  the  other  2  will  be  20 ;  if  the  first  1  is  called 
10,  then  2  will  be  20,  and  the  next  2  is  200,  etc.  When  the 
1  at  the  left  hand  is  taken  88. 1,  or  one  tenth,  the  line  is  read 
in  the  following  order:  1,  2,  3,  4,  5,  6,  7,  8,  9  tenths,  unity 
or  1 ;  2,  3,  4,  5,  .6,  7,  8,  9,  10;  when  a  higher  value  is  set 
on  them,  they  will  read  thus,  beginning  next  the  joint,  10,  20, 
30,  40,  50,  60,  70,  80,  90,  100,  200,  300,  and  so  on  to  1000. 


carpenters'  slide-rule.  299 

To  multiply  two  Numbers. 
Set  1  on  A  (1st  line)  to  either  of  the  given  numbers  on  B 
(2d  line),  then  at  the  other  number  on  A  (1st  line)  will  be 
found  the  product  on  B  (2d  line). 

Example. — Multiply  12  by  25. 

Under  1  on  A  put  12  on  B,  and  under  25  on  A  is  300  on  B. 

Ex:  2.  Multiply  64  by  15.  Ans.  960. 

Note. — If  the  third  terra  runs  beyond  the  end  of  the  line,  look  for  it 
on  the  first  radius  or  other  part  of  the  line,  and  increase  it  ten  times. 

To  divide  one  Number  by  another. 
Set  1  on  A  to  the  divisor  on  B,  then  at  the  dividend  on  B 
will  be  found  the  quotient  on  A. 

Example. — Divide  300  by  25. 

Under  1  on  A  put  25  on  B,  and  at  300  on  B  is  12  on  A. 

Note. — If  the  dividend  runs  beyond  the  end  of  the  line,  diminish  it 
10  or  100  times,  as  may  be  required,  to  make  it  fall  upon  A,  and  in- 
crease the  quotient  accordingly. 

To  ascertain  a  Fourth  Proportional. 
Set  the  first  term  upon  A  to  the  second  on  B,  then  at  the 
third  on  A  is  the  fourth  on  B. 

Example.  —  What  is  the  fourth  proportional  to  8,  20, 
and  30  "? 

Under  8  on  A  put  20  on  B,  then  at  30  on  A  is  75  on  B. 

To  ascertain  a  Mean  Proportional. 
Set  the  first  term  on  C  to  the  same  term  on  D,  then  at  the 
second  on  C  is  the  mean  on  D. 

Example. — What  is  the  mean  proportional  between  20 
and  80  ! 

Under  20  on  C  set  20  on  D,  and  at  80  on  C  is  40  on  D. 


'■j 


300  carpenters'  slide-rule. 

Bute  of  Three  Direct. 

In  the  Rule  of  Three  Direct,  there  are  three  numbers  given 
to  find  a  fourth,  that  shall  have  the  same  proportion  to  the 
third  as  the  second  has  to  the  first. 

The  operation  is,  as  the  first  term  upon  A  is  to  the  second 
upon  B,  so  is  the  third  term  upon  A  to  the  fourth  upon  B. 

Or,  bring  the  first  term  upon  B  to  the  second  upon  A ; 
then  against  the  third  upon  B  is  the  result  upon  A. 

Example. — If  a  man  can  walk  20  miles  in  5  hours,  how 
long  will  he  require  to  walk  125  miles  at  the  same  rate"? 

Over  20  upon  B  put  5  upon  A,  and  against  125  upon  B  is  31.25,  the 
result,  upon  A. 

Rule  of  Three  Inverse. 

In  this  rule,  there  are  three  numbers  given  to  find  a  fourth, 
that  shall  have  the  same  proportion  to  the  second  as  the  first 
has  to  the  third. 

Note. — If  more  requires  more,  or  less  requires  less,  the  question  he-+ 
longs  to  the  rule  of  three  Direct ;  but  if  more  requires  less,  or  less  re- 
quires more,  it  then  belongs  to  the  rule  of  three  Inverse. 

Example. — If  6  men  can  do  a  certain  piece  of  work  in  8 
days,  how  many  will  it  require  to  perform  the  same  in  3  days  ? 

Note. — In  inverse  proportion,  thex  slide  is  to  be  inverted  (by  with- 
drawing it  and  introducing  the  opposite  end  of  it) ;  then  the  question 
will  be  operated  in  the  same  way  as  in  direct  proportion. 

Invert  the  slide  in  the  groove,  and  over  8  upon  C  set  6  upon  A; 
then  at  3  upon  C  is  16,  the  result,  upon  A. 

Vulgar  and  Decimal  Fractions. 
To  reduce  a  Vulgar  Fraction  to  its  equivalent  Decimal  Expression. 

The  operation  is,  as  the  denominator  upon  A  is  to  1  upon 
B,  so  is  the  numerator  upon  A  to  the  decimal  required  upon  B. 

Example. — Reduce  i  to  a  decimal. 

Set  1  upon  B  to  4  upon  A;  then  at  1  upon  A  is  .25.,  the  result, 
upon  B. 


carpenters'  slide-rule.  301 

To  extract  the  Square  Root. 

When  the  lines  C  and  D  are  equal  at  both  ends,  C  is  a  ta- 
ble of  squares,  and  D  a  table  of  roots  ;  consequently,  opposite 
to  any  number  or  division  upon  C  is  its  square  root  upon  D. 

Example. — If  a  tower  30  feet  in  height  is  on  the  side  of  a 
river  which  is  40  feet  in  width,  what  must  be  the  length  of  a 
ladder  that  will  reach  from  the  opposite  side  of  the  river  to  the 
top  of  the  tower  % 

Operation. — Set  the  slide  even  at  both  ends,  and  over  30  upon  D  is  90 
on  C,  and  at  40  on  D  is  160  upon  C,  which,  when  added  to  90,  =250; 
then  under  250  upon  C  is  50  upon  D,  the  length  of  the  ladder. 

To  Square  a  Number. 

Set  1  upon  D  to  1  upon  C ;  then  against  the  number  upon 
D  will  be  found  the  square  upon  C. 

To  Cube  a  Number. 

Set  the  number  upon  C  to  1  or  10  upon  D,  and  against  the 
same  number  upon  D  will  be  its  cube  upon  C. 

Set  6  upon  C  to  10  upon  D,  and  at  6  upon  D  is  216  upon  C. 

Zand  Measuring. 

The  Gauge  points  for  measuring  land  are  the  number  of 
square  chains,  square  perches,  and  square  yards  that  are  con- 
tained in  an  acre.  If  the  dimensions  are  given  in  chains,  the 
gauge  point  is  1,  10, 100,  etc.,  upon  A ;  if  in  perches,  it  is  1G0  ; 
but  if  it  is  given  in  yards,  the  gauge  point  is  4840,  which  the 
length  upon  B  must  always  be  set  to,  and  opposite  the  breadth 
upon  A  is  the  result,  in  acres  and  parts,  upon  B. 

Example. — If  a  field  is  20  chains  50  links  in  length,  and 
4  chains  40  links  broad,  how  many  acres  does  it  contain  ? 

Set  20.5  upon  B  to  1  upon  A,  and  at  or  under  4.4  upon  A  is  9  upon 
B=*#e  result  in  acres. 


302  carpenters'  slide-rule. 

Mensuration  of  Solids. 

In  measuring  and  weighing  solid  bodies,  the  tables  of  Gauge 
points  upon  the  rule  are  always  to  be  made  use  of,  and  are 
thus  explained. 

1.  All  gauge  points  are  taken  on  the  line  A. 

2.  All  lengths  must  be  noted  on  the  line  B,  and  are  to  be 
set  to  the  gauge  point  on  the  line  A. 

3.  All  squares  and  diameters  are  found  on  the  line  D. 

4.  Opposite  the  square  or  diameter  on  the  line  D  is  the 
content  or  result  on  the  line  C ;  or,  as  the  length  upon  B  is 
to  the  gauge  point  upon  A,  so  is  the  square  or  diameter  upon 
D  to  the  content  upon  C. 

There  are  three  gauge  points  for  every  element  that  is  giv- 
en in  the  table  for  square ;  F.F.F.  signifying  that  when  the 
length  and  both  the  squares  are  feet,  the  gauge  point  is  to  be 
found  under  F.F.F.  in  the  same  line  with  that  of  the  element 
or  material  to  be  measured  or  weighed. 

If  the  length  is  given  in  feet,  and  both  the  squares  are 
inches,  then  the  gauge  point  is  under  F.I.I. ;  but  if  the  dimen- 
sions of  both  length  and  square  are  in  inches,  then  the  gauge 
point  is  under  I.I.I. 

There  are  two  gauge  points  for  every  thing  that  is  to  be 
measured  or  weighed  of  a  cylindric  form ;  first,  when  the 
length  is  in  feet  and  the  diameter  in  inches,  the  gauge  point  is 
under  F.I.  If  the  length  is  in  inches  and  the  diameter  in 
inches,  then  the  gauge  point  is  under  I.I. 

There  are  two  gauge  jjoints  to  weigh  or  measure  every  thing 
of  a  globular  figure. 

A  globe  having  but  one  dimension,  it  must  be  either  in  feet 
or  inches  ;  if  it  is  in  feet,  the  gauge  point  is  under  F. ;  if  it  is 
in  inches,  the  gauge  point  is  under  I. 

Note. — In  measuring  or  weighing  square  timber,  stone,  metals,  or 
any  other  bodies  that  are  unequal  sided,  a  mean  j>roportion  must  be  first 
estimated,  in  order  to  obtain  a  true  cube  or  square. 


carpenters'  slide-rule.  303 

The  general  rule  for  a  globe  is,  as  the  gauge  point  on  A  is 
to  the  diameter  on  B,  so  is  the  content  on  C  to  the  diameter 
on  D ;  or,  set  the  diameter  upon  B  to  the  gauge  point  upon 
A,  and  against  the  diameter  upon  D  is  the  result  upon  C. 

Example. — If  a  piece  of  timber  is  16  inches  broad,  6  inches 
thick,  and  20  feet  long,  how  many  cubic  feet  does  it  contain  ! 

Ascertain  the  mean  square  by  setting  16  upon  C  to  16  upon  D,  and 
opposite  to  6  upon  C  is  9.8  upon  D,  the  side  of  a  square  equal  to  16  by  6; 
having  thus  ascertained  the  true  square,  look  for  the  gauge  point  for 
cubic  feet,  and  under  E.I.I,  is  144 ;  set  20,  the  length,  upon  B  to  144 
upon  A,  and  against  9.8  upon  D  are  13.3  cubic  feet  upon  C. 

Round  timber  is  generally  measured  by  the  girth,  which 
is  ascertained  by  a  line  run  around  the  middle  of  the  tree  or 
log,  and  taking  one  fifth  of  the  girth  for  the  side  of  the  square. 
This  is  not  precisely  correct,  but  it  is  the  method  commonly 
practiced,  and  is  performed  on  this  rule,  after  the  girth  is 
taken,  as  follows. 

Example. — A  round  log  is  30  feet  in  length,  and  40  inches 
around  its  middle;  how  many  cubic  feet  does  it  contain? 

Here  one  fifth  of  the  girth  is  8  inches. 

Multiply  the  length,  30,  by  2,  and  set  the  result,  60,  upon  B  to  144 
upon  A,  and  against  8  upon  ~D  is  26.6,  the  result,  in  feet,  upon  C. 

The  circumference  of  the  above  log  being  40  inches,  the  diameter  is 
12.73  inches. 

Set  30,  the  length,  upon  B,  to  1833,  the  gauge  point,  upon  A,  and 
against  12.73  upon  D  are  26.5  feet,  the  result,  upon  C.  By  this  it  will 
be  seen  that  there  is  but  a  slight  difference  between  the  customary  and 
true  methods  of  measuring. 

Cylinder,  Globe,  and  Gone. 

Example. — If  a  cylinder  is  6  inches  long,  and  6  inches  in 
diameter,  how  many  cubic  inches  does  it  contain  ? 

The  gauge  point  for  cubic  inches  is  1273. 

Set  6  upon  B  to  1273  upon  A,  and  against  6  upon  D  are  169  cubic 
inches,  the  result,  upon  C. 


304  carpenters'  slide-rule. 


Cask  Gauging. 

The  gauging  of  casks  is  performed,  after  a  mean  diameter  is 
found,  exactly  in  the  same  manner  as  in  the  last  examples. 
Casks  are  generally  reduced  to  what  is  termed  four  Varieties; 
and  their  mean  diameters  may  be  found  by  multiplying  the  dif- 
erence  between  the  head  and  bung  diameters  of the  first  variety 
by  .7,  the  second  by  .63,  the  third  by  .50,  and  the  fourth  by  .52. 
The  respective  products  of  these  numbers,  added  to  the  head 
diameter,  will  give  the  mean  diameter. 

Set  the  length  upon  B  to  the  gauge  point  upon  A,  and 
over  the  mean  diameter  on  D  is  the  result  upon  C. 

Example. — In  a  cask  of  the  first  variety,  the  head  diameter 
is  24,  the  bung  diameter  28,  and  the  length  30  inches ;  how 
many  gallons  will  it  contain  % 

Set  30  upon  B  to  353,  the  gauge  point  for  the  imperial  measure,  upon 
A,  and  against  26.8,  the  mean  diameter,  upon  D  is  61,  the  resuk  in  gal- 
lons, upon  C. 

Ex.  2.  How  many  gallons  are  contained  in  a  cask  of  the 
second  variety,  the  head  diameter  18,  the  bung  diameter  23,  and 
length  28  inches  ? 

Set  28  upon  B  to  the  gauge  point  upon  A,  and  against  21.15,  the 
mean  diameter,  upon  D  is  35.2,  the  result  in  gallons,  upon  C. 

Ex.  3.  If  a  cask  of  the  third  variety  is  20.  inches  at  the  head, 
26  at  the  bung,  and  29  inches  long,  what  will  be  its  contents 
in  gallons  ? 

Set  29  upon  B  to  the  gauge  point  upon  A,  and  against  23.36,  the 
mean  diameter,  upon  D  is  44.5,  the  result  in  gallons,  upon  C. 

Ex.  4.  A  cask  of  the  fourth  variety  is  34  inches  long,  the 
head  diameter  26,  and  the  bung  diameter  32 ;  how  many 
gallons  will  it  hold  ? 

Set  34  upon  B  to  the  gauge  point  upon  A,  and  against  29.12,  the 
mean  diameter,  on  D  is  81.5,  the  result  in  gallons,  upon  C. 


carpenters'  slide-rule.  305 

Miscellaneous  Questions. 

Under  this  head  may  be  introduced  a  great  many  original 
questions,  as  well  as  such  as  could  not  be  introduced  in  regu- 
lar order  in  the  foregoing  rules. 

Of  a  Circle. 

The  Diameter  being  given,  to  ascertain  the  Area  ;  or  the  Area 
being  given,  to  ascertain  the  Diameter. 

Set  .7854,  the  area  of  unity,  upon  C  to  unity  or  10  upon 
D,  then  the  lines  C  and  D  will  be  a  table  of  areas  and  diam- 
eters ;  for  against  any  diameter  upon  D  is  the  area  in  square 
inches  upon  C. 

The  Circumference  being  given,  to  ascertain  the  Area  ;  or  the 
Area  being  given,  to  ascertain  the  Circumference. 

Set  .0795  upon  C  to  1  or  10  upon  D,  then  the  lines  C  and 
D  will  be  a  table  of  areas  and  circumferences ;  for  against  any 
circumference  upon  D  is  the  area  in  square  inches  upon  C. 

The  Circumference  being  given,  to  ascertain  the  Diameter;  or 
the  Diameter  being  given,  to  ascertain  the  Circumference. 

Set  1  upon  B  to  3.141  upon  A,  then  tfce  lines  A  and  B  will 
be  a  table  of  diameters  and  circumferences ;  for  against  any 
diameter  upon  B  is  the  circumference  upon  A. 

To  ascertain  the  Side  of  a  Square  equal  in  Area  to  any  given 

Circle. 
Set  .886  upon  B  to  1  upon  A,  then  against  any  diameter 
of  a  circle  upon  A  is  the  side  of  a  square  that  will  be  equal  in 
area  upon  B. 

To  ascertain  the  Side  of  the  greatest  Square  that  can  be  inscribed 
in  any  given  Circle. 
Set  .707  upon  B  to  1  upon  A,  and  against  any  diameter  of  a 
circle  upon  A  is  the  side  of  its  greatest  inscribed  square  upon  B. 


306 


carpenters'  slide-rule. 


To  ascertain  the  Side  of  the  greatest  Equilateral  Triangle  that  can 
be  inscribed  in  any  given  Circle. 
Set  1  upon  B  to  115  upon  A,  and  against  any  diameter 
of  a  circle  upon  A  is  the  length  of  a  side  of  a  triangle  upon  B. 

The  Weights  of  Metals,  and  various  other  Results,  may  be  ob- 
tained in  a  similar  manner,  for  the  rules  of  operation  of  which, 
reference  is  given  to  the  Book  of  Directions  usually  furnished 
with  the  slide-rule. 


„       0 


Illustrations. 
What  is  the  weight  of  a  piece  of  cast  iron  3  inches  square 
and  6  feet  long  ! 

The  cast  iron  gauge  point  for  feet  long  and  inches  square  is  32.  See 
the  rule. 

Set  6  upon  B  to  32  upon  A,  and  against  3  upon  D  is  168,  the  result, 
in  pounds,  upon  C. 

A  cylinder  is  6  inches  in  length  and  6  inches  in  diameter ; 
what  is  its  weight  in  cast  iron,  wrought  iron,  and  brass? 

1st.  For  cast  iron,  the  gauge  point  is  489. 

Set  6  upon  B  to  489  upon  A,  and  against  6  upon  D  is  44  upon  C. 

2d.  For  wrought  irori,  the  gauge  point  is  453. 

Set  6  upon  B  to  453  upon  A,  and  against  6  upon  D  is  47.5  upon  C. 

3d.  For  brass,  the  gauge  point  is  424. 

Set  6  upon  B  to  424  upon  A,  and  against  6  upon  D  is  511,  the  result, 
ur  on  C. 


TABLE   OF   ADDITIONAL   GAUGE   POINTS. 


Square. 

Cylinder. 

Globe. 

F.F.F. 

F.I.I. 

1. 1. 1. 

p.i. 

i.i. 

p. 

i.  1 

Oak 

174 
15 
155 
591 
795 
8 
193 

252 
217 
243 
85 
115 
115 
278 

303 
2605 
269 
102 
138 
138 
333 

320 
276 
31 
116 
147 
146 
354 

386 
333 
342 
13 
176 
126 
424 

332 
286 
296 
113 
152 
153 
369 

578 

Mahogany 

Box 

49 
512 
195 
263 
264 
637 

Marble 

Brick 

GAUGING.  307 


CASK  GAUGING. 

The  operation  of  cask  gauging  is  ordinarily  performed  with 
the  aid  of  five  instruments,  viz.,  a  Gauging  Slide-rule,  a"  Gauging 
or  Diagonal  Rod,  Callipers,  a  Bung  Rod,  and  a  Wantage  Rod. 

THE    GAUGING    SLIDE-RULE.* 

The  Gauging  Slide-rule  is  a.  flat  rule,  very  similar  to  an  or- 
dinary slide-rule,  except  that  it  is  not  jointed,  and  its  being 
adapted  for  use  for  the  purpose  of  measuring  and  gauging 
casks,  in  addition  to  those  of  the  ordinary  computations  effect- 
ed by  a  slide-rule. 

Upon  the  plain  or  outer  face  there  are  jive  lines;  the  first 
three  are  alike,  being  equally  divided,  and  all  *>f  the  same  ra- 
dius,! and  each  containing  twice  the  length  of  one. 

The  fourth  line  is  differently  divided  from  the  others,  and 
is  used  in  the  operation  of  gauging,  in  the  determination  of 
the  contents  of  casks  when  Lying,  by  the  element  of  the  depth 
of  liquor  within  them,  which  is  termed  the  wet  inches. 

The  fifth  line  is  similarly  divided  to  the  fourth,  and  is  used 
in  the  operation  of  gauging,  in  the  determination  of  the  con- 
tents of  casks,  when  Standing,  by  the  element  of  the  depth  of 
liquor  within  them,  which  is  also  termed  the  wet  inches. 

Note. — The  operation  of  gauging  in  this  manner — that  is,  by  the 
element  of  wet  inches — is  termed  Ullaging. 

Upon  the  opposite  or  inner  face  there  are  four  lines ;  the 
first  js  divided  to  represent  gallons,{  the  second  is  a  line  of 
mean  diameters,  and  the  third  and  fourth  lines  are  divided  into 
inches  and  tenths. 

*  As  manufactured  by  Belcher,  Brothers,  &  Co.,  New  York. 

t  The  first  three  lines  are  divided  alike  to  the  ordinary  carpenters' 
slide-rule,  or  Gunter's  line,  described  at  page  297,  and  the  operations 
of  multiplication,  division,  etc.,  etc.,  may  be  performed,  by  inspection, 
as  there  described.  » 

J  231  cubic  inches,  which  is  the  U.  S.  standard  gallon. 


308  GAUGING. 

The  third  line  is  divided  each  way  from  the  thumb-piece, 
running  from  38  to  G3  to  the  right,  and  from  0  to  12  to  the 
left. 

The  use  of  this  line  is  to  measure  the  diameter  of  the  head 
of  a  cask,  and  is  thus  operated : 

Place  the  outside  edge  of  the  brass  shoulder  on  the  right 
end'of  the  gauge,  on  the  head  of  a  cask,  and  close  to  the  inside 
of  a  stave  in  line  with  the  centre  of  the  head ;  move  the  thumb- 
piece  until  its  perpendicular  or  left  face  is  in  a  line  with  a 
point  at  the  other  end,  which  would  give  the  diameter  of  the 
head  on  its  inside ;  then  remove  the  gauge,  and  on  the  fourth 
line,  under  the  face  of  the  thumb-piece,  read  off  the  diameter 
of  the  head  in  inches  and  tenths. 

If  the  diameter  of  the  head  exceeds  38  inches,  then  it  is  to 
be  found  on  the  third  line,  at  the  left  end  of  the  gauge,  on  the 
line  running  from  38  to  63. 

On  the  left  of  the  thumb-piece  is  a  scale  of  12  inches  and 
tenths  of  inches,  and  above  it  is  a  Scale  of  1st  Varieties ;  that 
is,  varieties  of  the  first  form  of  casks,  the  use  of  which  is  here- 
after explained  under  the  head  of  Varieties. 

The  fourth  line  is  divided  into  inches  and  tenths,  running 
from  the  right  to  the  left,  and  is  used  for  the  purposes  of  or- 
dinary measurement. 

Upon  one  side  of  the  instrument  is  a  Scale  of  2d  Varieties, 
the  use  of  which,  and  of  all  like  scales,  is  to  obtain  by  inspec- 
tion the  mean  diameter  of  casks  of  the  different  varieties  of  fig- 
ure, and  which  are  thus  classed. 

Varieties  of  Casks.* 
First  Variety. — Casks  of  the  ordinary  form,  being  that  of 
the  Middle  Frustrum  of  a  Prolate  Spheroid,  as  Fig.l. 

*  The  basis  of  determination  of  a  scale  of  varieties  is  that  of  giving 
a  multiplier  whereby  the  mean  diameter  of  arcask  may  be  ascertained, 
and  the  operation  is  effected  as  shown  on  page  310. 


GAUGING. 


309 


Rum  puncheons  and  whiskey  barrels  are  fair  exponents  of  this  form, 
which  comprises  all  casks  having  a  spherical  outline  of  stave. 

Second  Variety. — Casks  of  the  form  ot  the  Middle  Frustrum 
of  a  Parabolic  Spindle,  as  Fig.  2. 
Fig.  2.  a 


Wine  casks  are  exponents  of  this  form,  which  comprises  all  casks  in 
which  the  curve  of  the  staves  quickens  slightly  at  the  bilge. 

Third  Variety. — Casks  of  the  form  of  the  Middle  Frustrum 
of  a  Paraboloid,  as  Fig.  3. 

Fig.  3.  a 


Brandy  casks  and  provision  barrels  are  exponents  of  this  form,  which 
comprises  all  casks  in  which  the  curve  of  the  staves  quickens  at  the 
chime. 


310 


GAUGING. 


Fourth  Variety. — Casks  of  the  form  of  two  equal  Frustrums 
of  Cones,  as  Fig.  4. 

Fig.  4.  a 


A  gin  pipe  is  an  exponent  of  this  form,  which  comprises  all  casks  in 
which  the  curve  of  the  staves  quickens  sharply  at  the  bilge. 

As  the  rule,  however,  is  provided  with  but  scales  of  two 
varieties,  it  is  usual  to  apply  the  first  scale  to  all  casks  in 
which  the  middle  diameter  (intermediate  between  the  bung 
and  head),  as  g  h,  Fig.  1,  approaches  nearest  to  that  of  the 
bung  diameter. 

The  scale  of  2d  variety  upon  the  edge  of  the  rule  is  adapt- 
ed to  all  casks  in  which  the  middle  diameter  approaches  next 
or  second  in  order  of  proportion  to  that  of  the  bung  diameter, 
as  Fig.  2. 

Scales  of  3d  and  4th  varieties  are  not  given.  Such  scales, 
however,  are  wanted,  and  are  applicable  to  all  casks  in  which 
the  middle  diameter  bears  a  less  proportion  to  the  bung  diam- 
eter than  in  any  of  the  other  varieties. 

%  To  ascertain  the  Mean  Diameter  of  a  Cash 

Kule. — Subtract  the  head  diameter  from  the  bung  diameter 
in  inches,  and  multiply  the  difference  by  the  following  units 
for  the  four  varieties ;  add  the  product  to  the  head  diameter, 
and  the  sum  will  give  the  mean  diameter  of  the  varieties  re- 
quired. 

1st  variety 7    I  3d  variety .56 

2d  variety 63  I  4th  variety 52 


GAUGING. 


311 


Example. — The  bung  and  head  diameters  of  a  cask  of  the 
1st  variety  are  24  and  20  inches  ;  what  is  its  mean  diameter? 

24  _  20  =  4,  and  4  X  .7  =  2.8,  which,  added  to  20,  =  22.8  inches,  the 
mean  diameter. 

Ex.  2.  The  bung  and  head  diameters  of  a  cask  of  the  2d 
variety  are  23  and  20  inches ;  what  is  its  mean  diameter  % 

Ans.  21.89  inches. 

Operation  by  the  Gauging  Slide-rule. 

Example  1.  Subtract  20  from  24,  and  over  4,  the  difference,  on  the 
line  of  inches  of  the  scale  of  1st  varieties,  read  2.76,  which,  added  to  20, 
=22.76,  the  result  required. 

Ex.  2.  Subtract  20  from  23,  and  under  3,  the  difference,  on  the  line 
of  inches  of  the  scale  of  2d  varieties,  read  1.88,  which,  added  to  20,  = 
21.88,  the  result  required. 


THE    GAUGING   OK   DIAGONAL   ROD. 

The  Gauging  or  Diagonal  Bod  is  a  square  rule  having  four 
faces,  being  commonly  four  feet  long.  This  instrument  is  used 
both  for  gauging  and  measuring  casks,  and  in  gauging,  the  con- 
tents are  ascertained  from  one  dimension  only,  viz.,  the  diagonal 
of  the  cask,  or  the  length  from  the  centre  of  the  bung-hole  to 
the  junction  of  the  head  of  the  cask  with  the  stave  opposite 
to  the  bung,  being  the  longest  straight  line  that  can  be  drawn 
within  a  cask  from  the  centre  of  the  bung.  Accordingly,  on 
two  opposite  faces  of  the  rule  are  scales  of  inches  for  measur- 
ing this  diagonal,  between  which,  on  a  third  face,  are  placed 
the  contents  in  gallons. 

To  ascertain  the  Contents  of  a  Cask  by  the  Gauging  or  Diagonal 
Bod,  Fig.  5. 
Fig.  5.  a 


I /  :  ' 

/  i 

/ 
<4r 4x  _j- 


312  GAUGING. 

Operation. — Introduce  the  pointed  end  of  the  rod  into  the 
bung-hole  of  a  cask  (the  plane  face  of  the  rod  being  upper- 
most) until  it  reaches  the  junction  of  the  head  and  bottom  of 
the  cask  at  its  lowest  point ;  then  adjust  the  upper  end  of  the 
rod,  so  that  the  under  side  (divided  into  gallons)  shall  be  in 
the  centre  of  the  hole  in  a  line  with  the  under  side  of  the 
bung  stave.  Observe  this .  point,  by  the  aid  of  the  divisions 
of  inches  and  tenths  on  either  of  the  sides  of  the  rod  ;  with- 
draw it,  and,  in  a  line  with  the  division  observed,  read  off  on 
the  under  side  of  the  rod  the  contents  of  the  cask  in  gallons. 

Illustration. — In  the  preceding  figure,  the  bung  diameter,  a  b,  is  24 
inches,  the  head  diameter,  ef,  is  20  inches,  and  the  half  length, /r, 
is  18  inches. 

Hence,  by  Geometry,  the  height,  a  r,  of  the  triangle  afr  is  24— 

24. 20 

— - — =22.     Consequently,  the  length =22,  and  the  base  =  18,  the  hy- 

pothenuse,  a  f,  =28.425.  Then,  by  inspection  of  the  rod,  it  will  be 
seen  that  in  a  line  with  28.425  inches  is  63,  the  number  of  gallons  the 
cask  contains. 


THE   CALLIPERS. 

The  Callipers  is  a  sliding  rule  adapted  to  project  over  the 
chimes  of  casks  to  measure  their  inner  length,  and  when  it  is 
adjusted  to  the  heads  of  a  cask,  the  inner  length  of  it  may  be 
read  off,  a  difference  of  2  inches,  being  an  allowance  of  1  inch 
for  the  thickness  of  each  head,  being  provided  for  in  the  divis- 
ions of  the  rule. 

Note. — When  the  thickness  of  the  heads  is  known  to  differ  from  an 
inch  each,  the  difference  above  or  less  than  an  inch,  as  the  case  may 
be,  is  to  be  subtracted  from  or  added  to  the  length  indicated  by  the  cal- 
lipers. 

THE   BUNG  ROD. 

The  Bung  Rod  is  alike  to  the  diagonal  rod  in  construction, 
and  is  a  rod  for  determining  the  inner  diameter  of  a  cask,  or 
the  wet  inches  therein ;  and  in  order  to  enable  the  divisions  to 
be  accurately  noted,  there  is  a  slide  running  around  the  rod, 
with  a  collar  upon  its  lower  end,  which  is  brought  to  bear 


GAUGING. 


313 


upon  the  under  side  of  the  stave  at  the  bung ;  the  rod,  with 
the  slide  retained  in  position,  is  then  removed,  and  the  divis- 
ions on  the  rod  read  off. 

Note. — It  is  customary  to  combine  this  instrument  with  the  diagonal 
rod,  the  Inches  on  the  latter  answering  all  the  purposes  of  the  measure- 
ment required,  and  the  slide  is  removed  or  adjusted  as  may  be  required. 


THE   WANTAGE   ROD. 

The  Wantage  Rod  is  a  scale  having  four  equal  sides,  upon 
which  are  divisions  adapted  to  the  many  varieties  of  vessels, 
as  barrels,  tierces,  and  hogsheads. 

The  use  of  this  instrument  is  to  obtain  by  inspection  the 
number  of  gallons  of  liquor  deficient  in  a  cask,  when  the  de- 
ficiency does  not  reach  to  an  extent  that  would  class  the  ves- 
sel as  an  ullage  cask. 

There  is  a  metal  collar  upon  one  of  its  sides,  which  is  intro- 
duced into  the  bung-hole  of  a  cask  until  its  upper  side  is.  at 
the  under  side  of  the  bung  stave ;  the  instrument  is  then  re- 
moved, and  the  wet  line  indicates  the  number  of  gallons  (un- 
der the  designation  on  the  rod,  of  the  cask  to  which  it  is  ap- 
plied) the  cask  is  deficient,  or  wants  of  being  full. 

To  ascertain  the  Contents  of  a  Cask  of  the  1st  Variety,  Fig.  6. 
Fig.  6.  a 


By  Mensuration. 

Rule. — To  twice  the  square  of  the  bung  diameter,  a  b,  add 
the  square  of  the  head  diameter,  e  f;  multiply  this  sum  by 
the  length,  c  d,  of  the  cask,  and  the  product  again  by  .2618, 

O 


314  GAUGING. 

and  it  will  give  the  contents  in  cubic  inches,  which,  being  di- 
vided by  231,  will  give  the  result  in  gallons.* 

Example. — The  bung  and  head  diameters  of  a  cask,  a  b  and 
ef,  are  24  and  20  inches,  and  the  length,  c  d,  36 ;  what  are 
its  contents  in  gallons  ? 

242  X  2  +  203  =  1552,  which  X  36  =  55872,  and  55872  X  .2618  = 
14627.2896,  which  -f-231  =  63.32,  the  gallons  required. 

Ex.  2.  The  bung  and  head  diameters  of  a  cask  are  36 
and  30  inches,  and  the  length  54 ;  what  are  its  contents  in 
gallons?  Ans.  213.7  gallons. 

By  the  Gauging  Slide-rule. 

Operation. — Subtract  the  head  diameter  from  the  bung  di- 
ameter, and  look  for  the  difference  on  the  lower  line  (inches) 
of  the  scale  of  1st  varieties,  and  above  it,  on  the  second  line, 
is  the  mean  difference,  which  is  to  be  added  to  the  head  diam- 
eter'for  the  mean  diameter. 

Thus,  24— 20  = 4  =  difference  of  diameters. 

Above  4,  on  1st  line  of  scale,  read  2.75,  which,  added  to  20,  =22.75, 
the  mean  diameter  required. 

Then,  set  the  left  end  of  the  slide  on  the  inner  face  of  the  rule  to  the 
length  of  the  cask  (36)  on  the  first  line ;  look  for  the  mean  diameter 
(22.75)  on  the  second  line  (or  top  line  of  slide),  and  above  it,  on  the  first 
line,  read  63.6,  which  is  the  capacity  of  the  cask  in  gallons. 

Ex.  2.  The  bung  and  head  diameters  of  a  cask  are  36  and 
30  inches,  and  the  length  54 ;  what  are  its  contents  in  gal- 
lons? 

The  difference  of  diameters  is  6,  which,  by  the  scale,  =4. 16. 

-.  Ans.  214.6  gallons. 

Ex.  3.  The  length  of  a  cask  is  51  inches,  and  its  bung  and 
head  diameters  31  and  26  inches;  what  are  its  contents  in 
gallons?  Ans.  150.6  gallons. 

*  Whenever  the  contents  are  required  in  bushels,  divide  by  2150.4. 


GAUGING. 


315 


To  ascertain  the  Contents  of  a  Cash  of  the  2d  Variety,  Fig.  7. 

Fig.  7.  a 


By  Mensuration. 

Rule. — To  the  square  of  a  head  diameter  add  double  the 
square  of  the  bung  diameter,  and  from  the  sum  subtract  -^ 
of  the  square  of  the  difference  of  the  diameters ;  then  multiply 
the  remainder  by  the  length,  and  the  product  again  by  .2618, 
which,  being  divided  by  231,  will  give  the  contents  in  gallons. 

Example. — The  bung  and  head  diameters  of  a  cask,  a  b  and 
ef,  are  24  and  18  inches,  and  the  length,  c  d,  36 ;  what  are 
its  contents  in  gallons'? 


182+242x 2  =  1476,  and  1476—^.  of  24-18  =1461.6,  which  X 36 
=  52617.6,  and  52617.6 X. 2618  =  13775.288,  which-^23 1=59.63,  the 
gallons  required. 

By  the  Gauging  Slide-rule. 

Operation. — Subtract  the  head  diameter  from  the  bung  di- 
ameter, and  look  for  the  difference  on  the  upper  line  (inches) 
of  the  scale  of  2d  varieties,  and  below  it,  on  the  second  line, 
is  the  mean  difference,  which  is  to  be  added  to  the  head  diam- 
eter for  the  mean  diameter. 

Thus,  24  — 18 =%=  difference  of  diameters. 

Below  6,  on  2d  line  of  scale  of  2d  varieties,  read  3.8,  which,  added 
to  18,  =21.8,  the  mean  diameter  required. 

Then,  set  the  left  end  of  the  slide  on  the  inner  face  of  the  rule  to  the 
length  of  the  cask  (36)  on  the  first  line;  look  for  the  .mean  diameter 
(21.8)  on  the  second  line  (or  top  line  of  slide),  and  above  it,  on  the  first 
line,  read  58.2,  which  is  the  capacity  of  the  cask  in  gallons. 


316 


GAUGING. 


Ex.  2.  The  bung  and  head  diameters  of  a  cask  are  36  and 
27  inches,  and  the  length  54 ;  what  are  its  contents  in  gallons? 
The  difference  of  diameters  is  9,  which,  by  the  scale,  =5.77. 

Ans.  197.5  gallons. 

Ex.  3.  The  length  of  a  cask  is  51  inches,  its  bung  and  head 
diameters  30  and  26  inches  ;  what  are  its  contents  in  gallons? 

Ans.  141.3  gallons. 


To  ascertain  the  Contents  of  a  Cask  of  the  3d  Variety,  Fig.  8. 
Fig.  8.  a 


By  Mensuration. 

Rule. — To  the  square  of  the  bung  diameter  add  the  square 
of  the  head  diameter;  multiply  the  sum  by  the  length,  and 
the  product  again  by  .3927,  which,  being  divided  by  231,  will 
give  the  contents  in  gallons. 

Example. — The  bung  and  head  diameters  of  a  cask,  a  b  and 
e  /,  are  24  and  20  inches,  and  the  length,  c  d,  36;  what  are 
its  contents  in  gallons? 

242+203x 36=35136,  which X. 3927 =1379 7.907,  and  13797.907-r- 
231=59.73,  the  result  required.    . 

By  the  Gauging  Slide-rule. 

Operation. — Subtract  the  head  diameter  from  the  bung  di- 
ameter, and  multiply  the  difference  by  the  unit  .56,  page  310, 
which  is  to  be  added  to  the  head  diameter  for  the  mean  diam- 
eter. 

Thus,  20-24=4,  which x. 56=2.24,  and  20  +  2.24=22.24,  the  mean 
diameter  required. 


GAUGING. 


317 


Then,  set  the  left  end  of  the  slide  on  the  inner  face  of  the  rule  to  the 
length  of  the  cask  (36)  on  the  first  line ;  look  for  the  mean  diameter 
(22.24)  on  the  second  line  (or  top  line  of  slide),  and  above  it,  on  the  first 
line,  read  60.16,  which  is  the  capacity  of  the  cask  in  gallons. 

Ex.  2.  The  bung  and  head  diameters  of  a  cask  are  36  and 
30  inches,  and  the  length  54 ;  what  are  its  contents  in  gallons  1 
The  difference  of  diameters  is  6,  which,  by  the  rule,  =3.36. 

Ans.  204.8  gallons. 

Ex.  3.  The  length  of  a  cask  is  51  inches,  its  bung  and  head 
diameters  31  and  26  inches ;  what  are  its  contents  in  gallons? 

Ans.  144.9  gallons. 
Ex.  4.  The  length  of  a  cask  is  50  inches,  and  its  bung  and 
head  diameters  30  and  25  inches ;  what  are  its  contents  1 

Ans.  131.4  gallons. 

To  ascertain  the  Contents  of  a  Cask  of  the  4th  Variety,  Fig.  9. 
Fig.  9.  a 


By  Mensuration. 

Rule. — Add  the  square  of  the  difference  of  the  diameters 
to  three  times  the  square  of  their  sum  ;  then  multiply  the  sum 
by  the  length,  and  the  product  again  by  .06566,  and  it  will 
give  the  contents  in  cubic  inches,  which,  being  divided  by  231, 
will  give  the  result  in  gallons.  ■ 

Example. — The  bung  and  head  diameters  of  a  cask,  a  b  and 
ef  are  24  and  16  inches,  and  the  length,  c  d,  36 ;  what  are 
its  contents  in  gallons  ? 


24-16  +  (24  +  16)2x3=4864,  which  X  36  =  175104,  and  175104  X 
.06566  =  11497.329,  whiclK-231  =49.77,  the  gallons  required. 


318  GAUGING. 

By  the  Gauging  Slide-rule. 

Operation. — Subtract  the  head  diameter  from  the  bung  di- 
ameter, and  multiply  the  difference  by  the  unit  .52,  page  310, 
which  is  to  be  added  to  the  head  diameter  for  the  mean  diam- 
eter. 

Thus,  24-16  =  8,  which X. 52  =4.16,  and  16+4.16=20.16,  the  mean 
diameter  required. 

Then,  set  the  left  end  of  the  slide  on  the  inner  face  of  the  rule  to  the 
length  of  the  cask  (36)  on  the  first  line :  look  for  the  mean  diameter 
(20.16)  on  the  Second  line  (or  top  line  of  slide),  and  above  it,  on  the  first 
line,  read  49.8,  which  are  the  contents  of  the  cask  in  gallons. 

Ex.  2.  The  bung  and  head  diameters  of  a  cask  are  36  and  24 
inches,  and  the  length  54 ;  what  are  its  contents  in  gallons? 

The  difference  of  diameters  is  12,  which,  by  the  rule,  =6.24. 

Ans.  168.5  gallons. 

Ex.  3.  The  length  of  a  cask  is  51  inches,  its  bung  and  head 

diameters  31  and  23  inches ;  what  are  its  contents  in  gallons 

and  in  bushels?  Ans.  128.2  gallons, 

,  128.2x231     10_,    __ 
and     ^^  An    —13.77  bushels, 
2150.42 

To  ascertain  the  Contents  of  a  Cask  when  the  Dimensions  are  less 
than  the  Divisions  on  the  Scale  as  numbered,  viz.,  for  a  Length 
of  less  than  25  inches,  and  a  Mean  Diameter  of  less  than  17.3 
inches. 

Operation. — Determine  the  mean  diameter;  then  double 
both  the  length  and  the  mean  diameter;  ascertain  the  con- 
tents for  those  dimensions,  and  divide  the  result  by  8. 

Note. — When  only  one  of  the  dimensions  is  doubled,  divide  the  re- 
sult by  4. 

Example. *-The  dimensions  of  a  barrel  of  the  2d  variety, 
having  bung  and  head  diameters  of  12  and  10  inches,  is  16 
inches  in  length  ;  what  are  its  contents  ? 

Mean  diameter  =  11.26  inches. 

Set  the  left  end  of  the  slide  to  32  (16x2)  on  the  1st  line,  and  over 
22.52  (11.26x2),  on  the  second  line,  is  55.36  on  the  1st  line,  which 
-j-8=6.92,  the  contents  of  the  barrel  in  gallons. 


GAUGING.  319 

ULLAGE    CASKS. 

To  ascertain  the  Contents  of  Ullage  Casks. 
When  a  cask  is  only  partly  filled,  it  is  termed  an  ullage  cash, 
and  is  considered  in  two  positions,  viz.,  as  lying  on  its  side, 
when  it  is  termed  a  Segment  Lying  (S.L.),  or  as  standing  on 
its  end,  when  it  is  termed  a  Segment  Standing  (S.S.). 

To  Ullage  a  Lying  Cask. 

By  Mensuration. 

Divide  the  wet  inches  by  the  bung  diameter ;  find  the  quo- 
tient in  the  column  of  versed  sines  in  the  table  of  circular  seg- 
ments, page  134,  a.id  take  out  its  corresponding  segment ; 
then  multiply  this  segment  by  the  capacity  of  the  cask  in  gal- 
lons, and  the  product  again  by  1.25  for  the  ullage  required. 

Example. — The  capacity  of  a  cask  is  90  gallons,  the  bung 
diameter  being  82  inches ;  what  are  its  contents,  at  8  inches 
depth? 

8-^32  =  .25,  the  tab.  seg.  of  which  is  .15355,  which X 90  =  13.81  £5, 
which  X  1.25  =  17.2744,  the  contents,  in  gallons. 

By  the  Gauging  Slide-rule. 

Operation. — On  the  plane  face, 

Set  the  bung  diameter  on  3d  line  to  100  at  the  right  hand 
on  4th  line,  and  on  this  line,  under  the  wet  inches  on  3d  line, 
take  off  the  number,  and  set  it  (by  moving  the  slide)  on  3d 
line  to  100  on  4th  line,  and  under  the  capacity  of  the  cask  on 
the  1st  line  read  the  contents  in  gallons  required. 

Thus,  set  32  on  3d  line  to  100  on  4th  line,  and  under  8  on  3d  line 
take  off  17.75  on  4th  line,  and  set  it  on  3d  line  at  100  on  4th  line  ;  then 
under  90  on  1st  line  read  16.85,  the  contents  in  gallons. 

Ex.  2.  The  capacity  of  a  cask  being  92  gallons,  and  the 
bung  diameter  32,  required  the  ullage  of  the  segment  when 
the  wet  inches  are  8.  Ane.  17.85  gallons. 

Ex.  3.  The  wet  inches  in  a  lying  cask  are  12  inches,  the 
bung  diameter  24,  and  the  capacity  of  the  cask  70  gallons ; 
what  are  the  contents  of  the  cask?  Ans.  35.2  gallons. 


320  GAUGING. 


To  Tillage  a  Standing  Cask. 

By  Mensuration. 

Add  together  the  square  of  the  diameter  at  the  surface  of 
the  liquor,  the  square  of  the  head  diameter,  and  the  square 
of  double  the  diameter  taken  in  the  middle  between  the  two ; 
then  multiply  the  sum  by  the  wet  inches  (length  between  the 
surface  and  nearest  end),  the  product  again  by  .1309,  and 
divide  by  231  for  the  result  in  gallons. 

Example. — The  diameter  at  the  surface  of  the  liquor  is  29 
inches,  the  head  diameter  of  the  cask  is  24,  the  diameter  taken 
in  the  middle  of  the  two  is  27,  and  the  depth  of  the  liquor,  or 
wet  inches,  is  20  ;  what  are  the  contents  of  the  cask  % 


292+242+27x2  =4333,  which  X  10=43330,  and  43330  X.1309- 
231=24.554,  the  result,  in  gallons. 


By  the  Gauging  Slide-rule. 

Operation. — On  the  plane  face, 

Set  the  length  on  3d  line  to  100  at  the  right  hand  on  5th 
line,  and  on  this  line,  under  the  wet  inches  on  3d  line,  take  off 
the  number,  and  set  it  (by  moving  the  slide)  on  3d  line  to  100 
on  5th  line,  and  under  the  capacity  of  the  cask  on  the  1st  line 
read  the  contents,  in  gallons,  required. 

The  capacity  of  the  cask  being  94.5  gallons,  and  the  length  35  inches. 

Thus,  set  35  on  3d  line  to  100  on  5th  line,  and  under  10  on  3d  line 
take  off  26.8  on  5th  line,  and  set  it  at  100  on  5th  line ;  then  under 
94.5  on  1st  line  read  25.3,  the  contents,  in  gallons. 

Ex.  2.  The  length  of  a  standing  cask  is  24  inches,  the  wet 
inches  12,  and  the  capacity  of  it  64  gallons ;  what  are  its 
contents  in  gallons'?  Ans.  32.28  gallons. 

Ex.  3.  The  length  of  a  standing  cask  is  24  inches,  the  wet 
inches  16,  and  the  capacity  of  it  65  gallons;  what  are  its 
contents  in  gallons?  Ans.  44.6  gallons. 


GAUGING.  321 


To  ascertain  the  Contents  of  a  Cask  by  four  Dimensions. 
Kule. — Add  together  the  squares  of  the  bung  and  head 
diameters,  and  the  square  of  double  the  diameter  taken  in  the 
middle  between  the  bung  and  head  ;  then  multiply  the  sum  by 
the  length  of  the  cask,  and  the  product  by  .1309  ;  then  divide 
this  product  by  231  for  the  result  in  gallons. 

Example. — What  are  the  contents  of  a  cask,  the  length  of 
which  is  40  inches,  the  bung  diameter  32,  the  head  diameter 
24,  and  the  middle  diameter  between  the  bung  and  the  head 
28.75  inches? 

322+242  =  1600=s«ro  of  squares  of  bung  and  head  diameters. 

28.75x2  =3306.25,  and  3306.25  +  1600  =  4906.25=sw7«  of  squares  of 
bung  and  head  diameters  and  of  double  the  middle  diameter. 

4906.25  X  40  =  \§&2b0=  product  of  above  sum  and  the  length  of  the  ca??: 
Tlien,  196250  X.  1309 =25689.125  =number  of  cubic  inches  in  the  casic, 
which-f-231  =  11 1.2083  gallons. 

Ex.  2.  The  bung  and  head  diameters  of  a  cask  are  24  and 
16  inches,  the  middle  diameter  20.5,  and  the  length  of  it  36 
inches ;  what  are  its  contents  in  gallons  ? 

Ans.  51.26  gallons. 

To  ascertain  the  Contents  of  any  Cash  from  three  Dimensions  only. 
Rule. — Add  into  one  sum  39  times  the  square  of  the  bung 
diameter,  25  times  the  square  of  the  head  diameter,  and  26 
times  the  product  of  the  two  diameters ;  then  multiply  the 
sum  by  the  length,  and  the  product  by  .008726  ;  then  divide 
the  quotient  by  231  for  the  result  in  gallons. 

Example. — The  length  of  a  cask  is  35.5  inches,  its  bung 
diameter  28.3  inches,  and  its  head  diameter  24  inches ;  what 
are  its  contents  in  gallons  ? 

28.32X  39  =31234.71  =39  times  the  square  of  the  bung  diameter. 

242  X  25  =  14400  =  25  times  the  square  of  the  head  diameter. 

28.3  X  24  X  26  =  17659.2=26  times  the  product  of  the  two  diameters. 

31234.71+14400  +  17659.2  =  63293.91,  which  x  35.5  =  2246933.8  = 
the  sum  of  the  above  products  X  the  length,  which  X  .008726=19606.744  = 
number  of  cubic  inches,  which-f-231  =84.88  —gallons  required. 
-       02 


322  GAUGING. 

Ex.  2.  What  are  the  contents  of  a  cask,  the  length  being 
40,  and  the  bung  and  head  diameters  32  and  24  inches? 

Ans.  112.273  gallons. 

Note. — This  is  the  most  exact  rule  of  any  for  three  dimensions,  and 
agrees  nearly  with  the  result  as  determined  by  a  diagonal  rod. 

Illustration. — If  a  diagonal  rod  was  applied  to  a  cask  of  the 
dimensions  above  given  (Ex.  2),  the  length  or  distance  determ- 
ined by  it  would  be  34.4. 

An  inspection  of  the  rod  will  show  110.1  gallons  to  be  in- 
dicated at  this  point. 

Note. — The  Divisor  for  English  ale  gallons  is  282,  and  for  Imperial 
gallons  277.274. 


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DEC  21 


fcEC'D  LD 


MAR  Z  9  laoO 


:  MAY  1 0 1970  12 

"      '      iiilMMI*  i&~ 


MAY    6f?Q 


|OAH  0EPAJTO*a« 


S^^^^i^ 


OCT  l  5  ZGOZ 


U.  C.  BERKELEY 


LD  21-100m-7,'33 


OA. 


ASK  LI 


H3S 


